smc-634-20-Log/Exp Function

Calculus, MET2 2023 VCE SM-Bank 1

The function \(g\) is defined as follows.

\(g:(0,7] \rightarrow R, g(x)=3\, \log _e(x)-x\)

Sketch the graph of \(g\) on the axes below. Label the vertical asymptote with its equation, and label any axial intercepts, stationary points and endpoints in coordinate form, correct to three decimal places. (3 marks)

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i. Find the equation of the tangent to the graph of \(g\) at the point where \(x=1\). (1 mark)

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ii. Sketch the graph of the tangent to the graph of \(g\) at \(x=1\) on the axes in part a. (1 mark)

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Newton's method is used to find an approximate \(x\)-intercept of \(g\), with an initial estimate of \(x_0=1\).

Find the value of \(x_1\). (1 mark)

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Find the horizontal distance between \(x_3\) and the closest \(x\)-intercept of \(g\), correct to four decimal places. (1 mark)

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i. Find the value of \(k\), where \(k>1\), such that an initial estimate of \(x_0=k\) gives the same value of \(x_1\) as found in part \(c\). Give your answer correct to three decimal places. (2 marks)

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ii. Using this value of \(k\), sketch the tangent to the graph of \(g\) at the point where \(x=k\) on the axes in part a. (1 mark)

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a.
b.i	\(y=2x-3\)
b.ii
c.	\(\dfrac{3}{2}=1.5\)
d.	\(0.0036\)
e.i	\(k=2.397\)
e.ii

Show Worked Solution

b.i	\(g(x)\)	\(=3\log_{e}x-x\)
	\(g(1)\)	\(=3\log_{e}1-1=-1\)
	\(g'(x)\)	\(=\dfrac{3}{x}-1\)
	\(g'(1)\)	\(=\dfrac{3}{1}-1=2\)

\(\text{Equation of tangent at }(1, -1)\ \text{with }m=2\)

\(y+1=2(x-1)\ \ \rightarrow \ \ y=2x-3\)

b.ii

c. \(\text{Newton’s Method}\)

\(x_1\)	\(=x_0-\dfrac{g(x)}{g'(x)}\)
	\(=1-\dfrac{-1}{2}\)
	\(=\dfrac{3}{2}=1.5\)

\(\text{Using CAS:}\)

d. \(\text{Using CAS}\)

\(x\text{-intercept}:\ x=1.85718\)

\(\therefore\ \text{Horizontal distance}=1.85718-1.85354=0.0036\)

e.i. \(\text{Using CAS}\)

\(k-\dfrac{3\log_{e}x-x}{\dfrac{3}{x}-1}\)	\(=1.5\)
\(k>1\ \therefore\ \ k\)	\(=2.397\)

e.ii

Calculus, MET2 2022 VCAA 3 MC

The gradient of the graph of `y=e^{3 x}` at the point where the graph crosses the vertical axis is equal to

0
`\frac{1}{e}`
1
`e`
3

Show Answers Only

`E`

Show Worked Solution

Graph crosses vertical axis when `x = 0`

`y`	`= e^(3x)`
`dy/dx`	`= 3e^(3x)`

When `x = 0`, `m = 3 xx e^(3 xx 0) = 3`

`=>E`

Calculus, MET2 2023 VCAA 3

Consider the function \(g:R \to R, g(x)=2^x+5\).

State the value of \(\lim\limits_{x\to -\infty} g(x)\). (1 mark)
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The derivative, \(g'(x)\), can be expressed in the form \(g'(x)=k\times 2^x\).
Find the real number \(k\). (1 mark)
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1. Let \(a\) be a real number. Find, in terms of \(a\), the equation of the tangent to \(g\) at the point \(\big(a, g(a)\big)\). (1 mark)
  
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1. Hence, or otherwise, find the equation of the tangent to \(g\) that passes through the origin, correct to three decimal places. (2 marks)
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Let \(h:R\to R, h(x)=2^x-x^2\).

Find the coordinates of the point of inflection for \(h\), correct to two decimal places. (1 mark)
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Find the largest interval of \(x\) values for which \(h\) is strictly decreasing.
Give your answer correct to two decimal places. (1 mark)
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Apply Newton's method, with an initial estimate of \(x_0=0\), to find an approximate \(x\)-intercept of \(h\).
Write the estimates \(x_1, x_2,\) and \(x_3\) in the table below, correct to three decimal places. (2 marks)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

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For the function \(h\), explain why a solution to the equation \(\log_e(2)\times (2^x)-2x=0\) should not be used as an initial estimate \(x_0\) in Newton's method. (1 mark)
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There is a positive real number \(n\) for which the function \(f(x)=n^x-x^n\) has a local minimum on the \(x\)-axis.
Find this value of \(n\). (2 marks)
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a. \(5\)

b. \(\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)

ci. \(y=2^a\ \log_{e}{(2)x}-(a\ \log_{e}{(2)}-1)\times2^a+5\)

\(\text{or}\ \ y=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

cii. \(y=4.255x\)

d. \((2.06 , -0.07)\)

e. \([0.49, 3.21]\)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} & -1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} & -0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} & -0.773 \\
\hline
\end{array}

g. \(\text{See worked solution.}\)

h. \(n=e\)

Show Worked Solution

a. \(\text{As }x\to -\infty,\ \ 2^x\to 0\)

\(\therefore\ 2^x+5\to 5\)

b.	\(g(x)\)	\(=2^x+5\)
		\(=\Big(e^{\log_{e}{2}}\Big)^x\)
		\(=e\ ^{x\log_{e}{2}}+5\)
	\(g'(x)\)	\(=\log_{e}{2}\times e\ ^{x\log_{e}{2}}\)
		\(=\log_{e}{2}\times 2^x\)
	\(\therefore\ k\)	\(=\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)

ci. \(\text{Tangent at}\ (a, g(a)):\)

\(y-(2^a+5)\)	\(=\log_{e}{2}\times2^a(x-a)\)
\(\therefore\ y\)	\(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

♦♦ Mean mark (c)(i) 50%.

cii. \(\text{Substitute }(0, 0)\ \text{into equation from c(i) to find}\ a\)

\( y\)	\(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\)	\(=2^a\ \log_{e}{(2)\times 0}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\)	\(=-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

\(\text{Solve for }a\text{ using CAS }\rightarrow\ a\approx 2.61784\dots\)

\(\text{Equation of tangent when }\ a\approx 2.6178\)

\( y\)	\(=2^{2.6178..}\ \log_{e}{(2)x}+0\)
\(\therefore\ y\)	\(=4.255x\)

♦♦♦ Mean mark (c)(ii) 30%.
MARKER’S COMMENT: Some students did not substitute (0, 0) into the correct equation or did not find the value of \(a\) and used \(a=0\).

d.	\(h(x)\)	\(=2^x-x^2\)
	\(h'(x)\)	\(=\log_{e}{(2)}\cdot 2^x-2x\ \ \text{(Using CAS)}\)
	\(h”(x)\)	\(=(\log_{e}{(2)})^2\cdot 2^x-2\ \ \text{(Using CAS)}\)

\(\text{Solving }h”(x)=0\ \text{using CAS }\rightarrow\ x\approx 2.05753\dots\)

\(\text{Substituting into }h(x)\ \rightarrow\ h(2.05753\dots)\approx-0.070703\dots\)

\(\therefore\ \text{Point of inflection at }(2.06 , -0.07)\ \text{ correct to 2 decimal places.}\)

e. \(\text{From graph (CAS), }h(x)\ \text{is strictly decreasing between the 2 turning points.}\)

\(\therefore\ \text{Largest interval includes endpoints and is given by }\rightarrow\ [0.49, 3.21]\)

♦♦ Mean mark (e) 40%.
MARKER’S COMMENT: Round brackets were often used which were incorrect as endpoints were included. Some responses showed the interval where the function was strictly increasing.

f. \(\text{Newton’s Method }\Rightarrow\ x_a-\dfrac{h(x_a)}{h'(x_a)}\)

\(\text{for }a=0, 1, 2, 3\ \text{given an initial estimation for }x_0=0\)

\(h(x)=2x-x^2\ \text{and }h'(x)=\ln{2}\times 2^x-2x\)

\begin{array} {|c|l|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} & 0-\dfrac{2^0-2\times 0}{\ln2\times 2^0\times 0}=-1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} & -1.433-\dfrac{2^{-1.433}-2\times -1.433}{\ln2\times 2^{-1.433}\times -1.433}=-0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} & -0.897-\dfrac{2^{-0.897}-2\times -0.897}{\ln2\times 2^{-0.897}\times -0.897}=-0.773 \\
\hline
\end{array}

g. \(\text{The denominator in Newton’s Method is}\ h'(x)=\log_{e}{(2)}\cdot 2^x-2x\)

\(\text{and the calculation will be undefined if }h'(x)=0\ \text{as the tangent lines are horizontal}.\)

\(\therefore\ \text{The solution to }h'(x)=0\ \text{cannot be used for }x_0.\)

♦♦♦ Mean mark (g) 20%.

h. \(\text{For a local minimum }f(x)=0\)

\(\rightarrow\ n^x-x^n=0\)

\(\rightarrow\ n^x=x^n\ \ \ (1)\)

\(\text{Also for a local minimum }f'(x)=0\)

\(\rightarrow\ \ln(n)\cdot n^x-nx^{n-1}=0\ \ \ (2)\)

\(\text{Substitute (1) into (2)}\)

\(\ln(n)\cdot x^n-nx^{n-1}=0\)

\(x^n\Big(\ln(n)-\dfrac{n}{x}\Big)=0\)

\(\therefore\ x^n=0\ \text{or }\ \ln(n)=\dfrac{n}{x}\)

\(x=0\ \text{or }x=\dfrac{n}{\ln(n)}\)

\(\therefore\ n=e\)

♦♦♦ Mean mark (h) 10%.
MARKER’S COMMENT: Many students showed that \(f'(x)=0\) but failed to couple it with \(f(x)=0\). Ensure exact values are given where indicated not approximations.

Calculus, MET2 2020 VCAA 4

The graph of the function `f(x)=2xe^((1-x^(2)))`, where `0 <= x <= 3`, is shown below.

Find the slope of the tangent to `f` at `x=1`. (1 mark)
Find the obtuse angle that the tangent to `f` at `x = 1` makes with the positive direction of the horizontal axis. Give your answer correct to the nearest degree. (1 mark)
Find the slope of the tangent to `f` at a point `x =p`. Give your answer in terms of `p`. (1 mark)
i. Find the value of `p` for which the tangent to `f` at `x=1` and the tangent to `f` at `x=p` are perpendicular to each other. Give your answer correct to three decimal places. (2 marks)
ii. Hence, find the coordinates of the point where the tangents to the graph of `f` at `x=1` and `x=p` intersect when they are perpendicular. Give your answer correct to two decimal places. (3 marks)

Two line segments connect the points `(0, f(0))` and `(3, f(3))` to a single point `Q(n, f(n))`, where `1 < n < 3`, as shown in the graph below.

i. The first line segment connects the point `(0, f(0))` and the point `Q(n, f(n))`, where `1 < n < 3`.
Find the equation of this line segment in terms of `n`. (1 mark)
ii. The second line segment connects the point `Q(n, f(n))` and the point `(3, f(3))`, where `1 < n < 3`.
Find the equation of this line segment in terms of `n`. (1 mark)
iii. Find the value of `n`, where `1 < n < 3`, if there are equal areas between the function `f` and each line segment.
Give your answer correct to three decimal places. (3 marks)

Show Answers Only

`-2`
`117^@`
`2(1-2p^(2))e^(1-p^(2))” or “(2e-4p^(2)e)e^(-p^(2))`
i. `0.655`
ii. `(0.80, 2.39)`
i. `y_1=2e^((1-n^2))x`
ii. `y_2=(2n e^((1-n^2)) – 6e^(-8))/(n-3) (x-3) + 6e^(-8)`
iii. `n= 1.088`

Show Worked Solution

a. `f(x)=2xe^((1-x^(2)))`

`f^{′}(1)=-2`

♦ Mean mark part (b) 37%.

b. `text{Solve:}\ tan theta = -2\ \ text{for}\ \ theta in (pi/2, pi)`

`theta = 117^@`

c. `text{Slope of tangent}\ = f^{′}(p)`

`f^{′}(p)=2(1-2p^(2))e^(1-p^(2))\ \ text{or}\ \ (2e-4p^(2)e)e^(-p^(2))`

d.i. `text{If tangents are perpendicular:}`

`f^{′}(p) xx -2 =-1\ \ =>\ \ f^^{′}(p)=1/2`

`text{Solve}\ \ 2(1-2p^(2))e^(1-p^(2))=1/2\ \ text{for}\ p:`

`p=0.655\ \ text{(to 3 d.p.)}`

d.ii. `text{Equation of tangent at}\ \ x=1: \ y=4-2x`

♦ Mean mark part (d)(ii) 41%.

`text{Equation of tangent at}\ \ x=p: \ y= x/2 + 1.991…`

`text{Solve}\ \ 4-2x = x/2 + 1.991…\ \ text{for}\ x:`

`=> x = 0.8035…`

`=> y=4-2(0.8035…) = 2.392…`

`:.\ text{T}text{angents intersect at (0.80, 2.39)}`

♦ Mean mark part (e)(i) 44%.

e.i. `Q (n, 2n e^(1-n^2))`

`m_(OQ) = (2n e^((1-n^2)) – 0)/(n-0) = 2e^((1-n^2))`

`:.\ text{Equation of segment:}\ \ y_1=2e^((1-n^2))x`

♦♦ Mean mark part (e)(ii) 28%.

e.ii. `P(3, f(3)) = (3, 6e^(-8))`

`m_(PQ) = (2n e^((1-n^2)) – 6e^(-8))/(n-3)`

`text{Equation of line segment:}`

`y_2-6e^(-8)`	`=(2n e^((1-n^2)) – 6e^(-8))/(n-3) (x-3)`
`y_2`	`=(2n e^((1-n^2)) – 6e^(-8))/(n-3) (x-3) + 6e^(-8)`

♦♦ Mean mark part (e)(iii) 28%.

e.iii. `text{Find}\ n\ text{where shaded areas are equal.}`

`text{Solve}\ int_(0)^(n)(f(x)-y_(1))\ dx=int_(n)^(3)(y_(2)-f(x))\ dx\ \ text{for} n:`

`=> n= 1.088\ \ text{(to 3 d.p.)}`

Calculus, MET2 2021 VCAA 3

Let `q(x) = log_e (x^2 - 1) - log_e (1 - x)`.

State the maximal domain and the range of `q`. (2 marks)
i. Find the equation of the tangent to the graph of `q` when `x = –2`. (1 mark)
ii. Find the equation of the line that is perpendicular to the graph of `q` when `x = –2` and passes through the point (–2, 0). (1 mark)

Let `p(x) = e^{-2x} - 2e^-x + 1.`

Explain why `p` is not a one-to one function. (1 mark)
Find the gradient of the tangent to the graph of `p` at `x = a`. (1 mark)

The diagram below shows parts of the graph of `p` and the line `y = x + 2`.

The line `y = x + 2` and the tangent to the graph of `p` at `x = a` intersect with an acute angle of `theta` between them.

Find the value(s) of `a` for which `theta = 60^@`. Give your answer(s) correct to two decimal places. (3 marks)
Find the `x`-coordinate of the point of intersection between the line `y = x + 2` and the graph of `p`, and hence find the area bounded by `y = x + 2`, the graph of `p` and the `x`-axis, both correct to three decimal places. (3 marks)

Show Answers Only

`text{Domain:} \ x ∈ (-∞, -1)`
`text{Range:} \ y ∈ R`
i. `-x – 2`
ii. `x + 2`
`text{Not a one-to-one function as it fails the horizontal line test.}`
`-2e^{-2a} + 2e^{-a}`
`-0.67`
`1.038\ text(u)^2`

Show Worked Solution

a. `text(Method 1)`

`x^2 – 1 > 0 \ \ => \ \ x > 1 \ \ ∪ \ \ x < -1`

`1 – x > 0 \ \ => \ \ x < 1`

`:. \ x ∈ (-∞, -1)`

`text(Method 2)`

`text{Sketch graph by CAS}`

`text{Asymptote at} \ x = -1`

`text{Domain:} \ x ∈ (-∞, -1)`

`text{Range:} \ y ∈ R`

b. i. `text{By CAS (tanLine} \ (q (x), x, –2)):`

`y = -x – 2`

ii. `text{By CAS (normal} \ (q (x), x, –2)):`

`y = x + 2`

c. `text{Sketch graph by CAS.}`

`p(x) \ text{is not a one-to-one function as it fails the horizontal line test}`

`text{(i.e. it is a many-to-one function)}`

d. `p′(x) = -2e^{-2x} + 2e^-x`

`p′(a) = -2e^{-2a} + 2e^{-a}`

`text{Case 1}`

`text{By CAS, solve:}`

`2e^{-a} -2e^{-2a} = – tan (15^@) \ \ text{for}\ a:`

`a = -0.11`

`text{Case 2}`

`text{Case 1}`

`text{By CAS, solve:}`

`2e^{-a} – 2e^{-2a} = -tan 75^@\ \ text{for}\ a:`

`a = -0.67`

f. `text{At intersection,}`

`x + 2 = e^{-2x} -2e^{-x} + 1`

`x = -0.750`

`text{Area}`	`= int_{-2}^{-0.750} x + 2\ dx + int_{-0.750}^0 e^{-2x} – 2e^{-x} + 1\ dx`
	`= 1.038 \ text(u)^2`

Calculus, MET2 2018 VCAA 9 MC

A tangent to the graph of `y = log_e(2x)` has a gradient of 2.

This tangent will cross the `y`-axis at

A. `0`

B. `-0.5`

C. `-1`

D. `-1 - log_e(2)`

E. `-2 log_e(2)`

Show Answers Only

`C`

Show Worked Solution

`y`	`= ln 2x`
`(dy)/(dx)`	`= 1/x`

`text(When)\ \ (dy)/(dx) = 2:`

`1/x`	`= 2`
`x`	`= 1/2`

`text(When)\ \ x = 1/2,\ \ y = ln 1 = 0`

`text(Find equation)\ \ m = 2,\ \ text(through)\ \ (1/2, 0):`

`y – 0`	`= 2 (x – 1/2)`
`y`	`= 2x – 1`

`=> C`

Algebra, MET2 2017 VCAA 4

Let `f : R → R :\ f (x) = 2^(x + 1)-2`. Part of the graph of `f` is shown below.

The transformation `T: R^2 -> R^2, \ T([(x),(y)]) = [(x),(y)] + [(c),(d)]` maps the graph of `y = 2^x` onto the graph of `f`.

State the values of `c` and `d`. (2 marks)
Find the rule and domain for `f^(−1)`, the inverse function of `f`. (2 marks)
Find the area bounded by the graphs of `f` and `f^(−1)`. (3 marks)
Part of the graphs of `f` and `f^(−1)` are shown below.

Find the gradient of `f` and the gradient of `f^(−1)` at `x = 0`. (2 marks)

The functions of `g_k`, where `k ∈ R^+`, are defined with domain `R` such that `g_k(x) = 2e^(kx)-2`.

Find the value of `k` such that `g_k(x) = f(x)`. (1 mark)
Find the rule for the inverse functions `g_k^(−1)` of `g_k`, where `k ∈ R^+`. (1 mark)
1. Describe the transformation that maps the graph of `g_1` onto the graph of `g_k`. (1 mark)
2. Describe the transformation that maps the graph of `g_1^(–1)` onto the graph of `g_k^(–1)`. (1 mark)
The lines `L_1` and `L_2` are the tangents at the origin to the graphs of `g_k` and `g_k^(–1)` respectively.
Find the value(s) of `k` for which the angle between `L_1` and `L_2` is 30°. (2 marks)
Let `p` be the value of `k` for which `g_k(x) = g_k^(−1)(x)` has only one solution.
i. Find `p`. (2 marks)
ii. Let `A(k)` be the area bounded by the graphs of `g_k` and `g_k^(–1)` for all `k > p`.
State the smallest value of `b` such that `A(k) < b`. (1 mark)

Show Answers Only

a. `c = −1, \ d = −2`

b. `f^(−1)(x) = log_2 (x + 2)-1, \ x ∈ (−2,∞)`

c. `3-2/(log_e(2))\ text(units)²`

d. `f′(0)= 2log_e(2) and f^(-1′)(0)= 1/(2log_e(2))`

e. `k = log_e(2)`

f. `g_k^(−1)(x)== 1/k log_e((x + 2)/2)`

g.i. `text(Dilation by factor of)\ 1/k\ text(from the)\ ytext(-axis)`

g.ii. `text(Dilation by factor of)\ 1/k\ text(from the)\ xtext(-axis)`

h. `k=sqrt3/6\ or\ sqrt3/2`

i.i. `p = 1/2`

i.ii. `b=4`

Show Worked Solution

a. `text(Using the matrix transformation:)`

`x′`	`=x+c\ \ => x=x′-c`
`y′`	`=y+d\ \ =>y= y′-d`

`y′-d`	`=2^(x′-c)`
`y′`	`= 2^(x′-c)+d`

`:. c = −1, \ d = −2`

b. `text(Let)\ \ y = f(x)`

`text(Inverse : swap)\ x\ ↔ \ y`

`x`	`= 2^(y + 1)-2`
`x + 2`	`= 2^(y + 1)`
`y + 1`	`= log_2(x + 2)`
`y`	`= log_2(x + 2)-1`

`text(dom)(f^(−1)) = text(ran)(f)`

`:. f^(−1)(x) = log_2 (x + 2)-1, \ x ∈ (−2,∞)`

c. `text(Intersection points occur when)\ \ f(x)=f^(−1)(x)`

MARKER’S COMMENT: Specifically recommends using `f^(-1)(x)-f(x)` in this type of integral to avoid errors in stating the integral.

`x`

`= −1, 0`

`text(Area)`	`= int_(−1)^0 (f^(−1)(x)-f(x))\ dx`
	`= 3-2/(log_e(2))\ text(units)²`

d.	`f′(0)`	`= 2log_e(2)`
	`f^(-1′)(0)`	`= 1/(2log_e(2))`

e. `g_k(x) = 2e^(kx)-2`

`text(Solve:)\ \ g_k(x) = f(x)quad text(for)\ k ∈ R^+`

`:. k = log_e(2)`

f. `text(Let)\ \ y = g_k(x)`

`text(Inverse : swap)\ x\ text(and)\ y`

`x`	`= 2e^(ky)-2`
`e^(ky)`	`=(x+2)/2`
`ky`	`=log_e((x+2)/2)`
`:. g_k^(−1)(x)`	`= 1/k log_e((x + 2)/2)`

♦♦ Mean mark part (g)(i) 31%.

g.i.	`g_1(x)`	`= 2e^x-2`
	`g_k(x)`	`= 2e^(kx)-2`

`:. text(Dilation by factor of)\ 1/k\ text(from the)\ ytext(-axis)`

♦♦ Mean mark part (g)(ii) 30%.

g.ii.	`g_1^(−1)(x)`	`= log_e((x + 2)/2)`
	`g_k^(−1)(x)`	`= 1/klog_e((x + 2)/2)`

`:. text(Dilation by factor of)\ 1/k\ text(from the)\ xtext(-axis)`

h. `text(When)\ \ x=0,`

♦♦♦ Mean mark part (h) 13%.

`g_k′(0)=2k\ \ => m_(L_1)=2k`

`g_k^(-1′)(0)=1/(2k)\ \ => m_(L_2)=1/(2k)`

`text(Using)\ \ tan 30^@=|(m_1-m_2)/(1+m_1m_2)|,`

`text(Solve:)\ \ 1/sqrt3`

`=+- ((2k-1/(2k)))/2\ \ text(for)\ k`

`:. k=sqrt3/6\ or\ sqrt3/2`

i.i `text(By inspection, graphs will touch once if at)\ \ x=0,`

♦♦♦ Mean mark part (i)(i) 4%.

`m_(L_1)`	`=m_(L_2)`
`2k`	`=1/(2k)`
`k`	`=1/2,\ \ (k>0)`

`:. p = 1/2`

i.ii `text(As)\ k→oo, text(the graph of)\ g_k\ text(approaches)\ \ x=0`

♦♦♦ Mean mark part (i)(ii) 2%.

`text{(vertically) and}\ \ y=–2\ \ text{(horizontally).}`

`text(Similarly,)\ \ g_k^(-1)\ \ text(approaches)\ \ x=–2`

`text{(vertically) and}\ \ y=0\ \ text{(horizontally).}`

`:. lim_(k→oo) A(k) = 4`

`:.b=4`

Calculus, MET1 SM-Bank 3

For the function `f:R→R,\ \ f(x)= 2e^x + 3x`, determine the coordinates of the point `P` at which the tangent to `f(x)` is parallel to the line `y = 5x - 3`. (3 marks)

Show Answers Only

`(0, 2)`

Show Worked Solution

`y = 5x − 3\ \ =>\ m=5`

`y`	`= 2e^x + 3x`
`(dy)/(dx)`	`= 2e^x + 3`

`text(Find)\ \ x\ \ text(when)\ \ (dy)/(dx) = 5,`

`5`	`= 2e^x + 3`
`2e^x`	`= 2`
`e^x`	`= 1`
`x`	`= 0`

`text(When)\ \ x = 0,`

`y`	`= 2e^0 + (3 xx 0)=2`

`:.P\ \ text{has coordinates (0, 2)}`

Calculus, MET1 SM-Bank 2

Let `f: (0,oo) → R,` where `f(x) = log_e (x).`

Find the equation of the tangent to `f(x)` at the point `(e, 1)`. (2 marks)

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`y = x/e`

Show Worked Solution

`y = log_ex`

`dy/dx = 1/x`

`text(At)\ (e, 1),\ \ m = 1/e`

`text(Equation of tangent,)\ \ m = 1/e,\ text(through)\ (e, 1):`

`y − 1`	`= 1/e(x −e)`
`:. y`	`= x/e`

Calculus, MET1 2007 VCAA 9

The graph of `f: R -> R`, `f(x) = e^(x/2) + 1` is shown. The normal to the graph of `f` where it crosses the `y`-axis is also shown.

Find the equation of the normal to the graph of `f` where it crosses the `y`-axis. (2 marks)
Find the exact area of the shaded region. (3 marks)

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`y = – 2x + 2`
`(2sqrte – 2)\ text(u)²`

Show Worked Solution

a. `text(Normal is)\ ⊥\ text(to tangent)`

MARKER’S COMMENT: A common error was made in calculating the point of tangency. Be careful!

`text(Point of tangency:)\ (0,2)`

`text(Gradient of normal:)`

`f′(x)`	`= 1/2 e^(x/2)`
`f′(0)`	`= 1/2`

`:. m_text(norm) = -2`

`text(Equation of normal:)`

`y – y_1`	`= m(x – x_1)`
`y – 2`	`= -2(x – 0)`
`y`	`=-2x+2`

♦ Mean mark 44%.

b.	`:.\ text(Area)`	`= int_0^1 (e^(x/2) + 1 – (-2x +2))\ dx`
		`= int_0^1 (e^(x/2) + 2x – 1)\ dx`
		`= [2e^(x/2) + x^2 – x]_0^1`
		`= (2e^(1/2) + 1^2 – 1) – (2e^0)`
		`= (2sqrte – 2)\ text(u)²`

Calculus, MET1 2009 VCAA 8

Let `f: R -> R,\ f(x) = e^x + k`, where `k` is a real number. The tangent to the graph of `f` at the point where `x = a` passes through the point `(0, 0).` Find the value of `k` in terms of `a.` (3 marks)

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`:. k = ae^a – e^a`

Show Worked Solution

`text(Point of tangency:)\ \ P (a, e^a + k)`

♦ Mean mark 45%.

`text(Find gradient of tangent at)\ \ x = a:`

`f(x)`	`=e^x+k`
`f′ (x)`	`= e^x`
`:. m_tan`	`= e^a`

`text(Find equation of tangent:)`

`y – y_1`	`= m (x – x_1)`
`y – (e^a + k)`	`= e^a (x – a)`

`text(Substitute)\ \ (0, 0):`

`0 – e^a – k`	`= – ae^a`
`:. k`	`= ae^a – e^a`
	`=e^a(a-1)`

Calculus, MET1 2012 VCAA 10

Let `f: R -> R,\ f(x) = e^(– mx) + 3x`, where `m` is a positive rational number.

i. Find, in terms of `m`, the `x`-coordinate of the stationary point of the graph of `y = f(x)`. (2 marks)
ii. State the values of `m` such that the `x`-coordinate of this stationary point is a positive number. (1 mark)
For a particular value of `m`, the tangent to the graph of `y = f(x)` at `x = – 6` passes through the origin.
Find this value of `m`. (3 marks)

Show Answers Only

i. `1/m log_e(m/3)`
ii. `m > 3`
`1/6`

Show Worked Solution

a.i. `text(SP’s occur when)\ \ f prime(x)=0`

`– me^(- mx) + 3`	`= 0`
`me^(-mx)`	`3`
`– mx`	`= log_e (3/m)`
`:. x`	`= – 1/m log_e (3/m)`
	`= 1/m log_e (m/3), \ m>0`

♦♦♦ Parrt (a)(ii) mean mark 18%.

ii.	`1/m log_e (m/3)`	`> 0`
	`log_e (m/3)`	`> 0`
	`m/3`	`> 1`
	`:. m`	`> 3`

b. `text(Point of tangency:)\ \ (– 6, e^(– 6m) – 18)`

♦♦ Mean mark 33%.
MARKER’S COMMENT: Many confused `m` with the more common use of `m` for gradient in `y=mx+c`.

`text(At)\ \ x=-6,`

`m_tan`	`= f prime (– 6)`
	`= – me^(– 6m) + 3`

`:.\ text(Equation of tangent:)`

`y – y_1`	`= m (x – x_1)`
`y – (e^(– 6m) – 18)`	`= (– me^(– 6m) + 3) (x – (– 6))`

`text(S)text{ince tangent passes through (0, 0):}`

`– e^(– 6m) + 18`	`= (–me^(– 6m) + 3)(6)`
`– e^(– 6m) + 18`	`= – 6 me^(– 6m) + 18`
`e^(– 6m) – 6me^(– 6m)`	`= 0`
`e^(– 6m) (1 – 6m)`	`= 0`
`1-6m`	`=0`
`:.m`	`=1/6`

Calculus, MET2 2012 VCAA 18 MC

The tangent to the graph of `y = log_e(x)` at the point `(a, log_e(a))` crosses the `x`-axis at the point `(b, 0)`, where `b < 0.`

Which of the following is false?

A. `1 < a < e`

B. The gradient of the tangent is positive

C. `a > e`

D. The gradient of the tangent is `1/a`

E. `a > 0`

Show Answers Only

`A`

Show Worked Solution

`text(Consider Option)\ A:`

♦♦ Mean mark 30%.

`text(T)text{angent at}\ x=1\ text{(from graph),}\ \ b>0.`

`text(T)text{angent at}\ \ x=e\ \ text{is}\ \ y=1/e x,\ \ b=0.`

`text(Graphing the tangents:)`

`text(For negative)\ xtext{-intercept (i.e.}\ \ b<0text{)}`

`a > e`

`:. A\ text(is false.)`

`=> A`

Calculus, MET2 2013 VCAA 11 MC

If the tangent to the graph of ` y = e^(ax), a != 0,\ \ text(at)\ \ x = c` passes through the origin, then `c` is equal to

A. `0`

B. `1/a`

C. `1`

D. `a`

E. `-1/a`

Show Answers Only

`B`

Show Worked Solution

`y = e^(ax)\ \ \ => dy/dx = ae^(ax)`

♦ Mean mark 47%.

`text(At)\ \ x = c,`

`y = e^(ac)\ \ \ => (dy)/(dx) = ae^(ac)`

`text(T)text(angent passes through)\ \ (0,0) and (c,e^(ac))`

`:.\ text(Gradient of tangent) = (e^(ac)-0)/(c-0) = e^(ac)/c`

`text(Equating the gradients,)`

`ae^(ac)`	`=e^(ac)/c`
`ac`	`=1`
`c`	`=1/a`

`=> B`