smc-5204-70-Sketch graph

Calculus, MET1 SM-Bank 21

The rule for function `f` is `f(x) = e^(-x^2)`. The diagram shows the graph `y = f(x)`.

The graph has two points of inflection.

Find the `x` coordinates of these points. (2 marks)
Explain why the domain of `f(x)` must be restricted if `f(x)` is to have an inverse function. (1 mark)
Find the rule for the inverse function `f^(-1)` if the domain of `f(x)` is restricted to `x ≥ 0`. (2 marks)
Find the domain for `f^(-1)`. (1 mark)
Sketch the curve `y = f^(-1) (x)`. (1 mark)

Show Answers Only

`x = +- 1/sqrt2`
`text(There can only be 1 value of)\ y\ text(for each value of)\ x.`
`f^(-1)x = sqrt(ln(1/x))`
`0 <= x <= 1`

Show Worked Solution

a.	`y`	`= e^(-x^2)`
	`dy/dx`	`= -2x * e^(-x^2)`
	`(d^2y)/(dx^2)`	`= -2x (-2x * e^(-x^2)) + e ^(-x^2) (-2)`
		`= 4x^2 e^(-x^2)-2e^(-x^2)`
		`= 2e^(-x^2) (2x^2-1)`

`text(P.I. when)\ \ (d^2y)/(dx^2) = 0`

`2e^(-x^2) (2x^2-1)`	`= 0`
`2x^2-1`	`= 0`
`x^2`	`= 1/2`
`x`	`= +- 1/sqrt2`

COMMENT: It is also valid to show that `f(x)` is an even function and if a P.I. exists at `x=a`, there must be another P.I. at `x=–a`.

`text(When)\ \ `	`x < 1/sqrt2,`	`\ (d^2y)/(dx^2) < 0`
	`x > 1/sqrt2,`	`\ (d^2y)/(dx^2) > 0`

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = 1/sqrt2`

`text(When)\ \ `	`x <-1/sqrt2,`	`\ (d^2y)/(dx^2) > 0`
	`x >-1/sqrt2,`	`\ (d^2y)/(dx^2) < 0`

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x =-1/sqrt2`

b.	`text(In)\ f(x), text(there are 2 values of)\ y\ text(for)`
	`text(each value of)\ x.`
	`:.\ text(The domain of)\ f(x)\ text(must be restricted)`
	`text(for)\ \ f^(-1) (x)\ text(to exist).`

c. `y = e^(-x^2)`

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

`x`	`= e^(-y^2),\ \ \ x >= 0`
`lnx`	`= ln e^(-y^2)`
`-y^2`	`= lnx`
`y^2`	`= -lnx`
	`=ln(1/x)`
`y`	`= +- sqrt(ln (1/x))`

`text(Restricting)\ \ x>=0,\ \ =>y>=0`

`:. f^(-1) (x)=sqrt(ln (1/x))`

d.	`f(0) = e^0 = 1`

`:.\ text(Range of)\ \ f(x)\ \ text(is)\ \ 0 < y <= 1`

`:.\ text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ 0 < x <= 1`

Functions, MET1 2008 VCAA 10

Let `f: R -> R,\ \ f(x) = e^(2x)-1`.

Find the rule and domain of the inverse function `f^(−1)`. (2 marks)
On the axes provided, sketch the graph of `y = f(f^(−1)(x))` for its maximal domain. (1 mark)
Find `f(−f^(−1)(2x))` in the form `(ax)/(bx + c)` where `a`, `b` and `c` are real constants. (2 marks)

Show Answers Only

`f^(−1)(x) = 1/2log_e(x + 1), x ∈ (−1,∞)`
`f(−f^(−1)(2x)) = (−2x)/(2x + 1)`

Show Worked Solution

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

`text(For Inverse, swap)\ x ↔ y`

`x`	`= e^(2y)-1`
`x + 1`	`= e^(2y)`
`2y`	`= log_e(x + 1)`
`y`	`= 1/2 log_e(x + 1)`

`text(Domain)(f^(−1))`	`=\ text(Range)\ (f)`
	`= (−1,∞)`

`:. f^(−1)(x) = 1/2log_e(x + 1),quadx ∈ (−1,∞)`

b. `f(f^(−1)(x)) = x`

♦ Mean mark part (b) 19%.
MARKER’S COMMENT: Few students were aware that a function of its own inverse function is the line `y=x` over the appropriate domain.

`text(Domain is)\ \ (-1, oo)`

♦ Mean mark 34%.

c.	`−f^(−1)(2x)`	`= −1/2 ln(2x + 1)`
	`:. f(−f^(−1)(2x))`	`= e^(-log_e(2x + 1))-1`
		`=(2x+1)^-1 -1`
		`= 1/(2x + 1)-1`
		`= (−2x)/(2x + 1)`

Calculus, MET2 2010 VCAA 1

Part of the graph of the function `g: (-4, oo) -> R,\ g(x) = 2 log_e (x + 4) + 1` is shown on the axes below
1. Find the rule and domain of `g^-1`, the inverse function of `g`. (3 marks)
2. On the set of axes above sketch the graph of `g^-1`. Label the axes intercepts with their exact values. (3 marks)
3. Find the values of `x`, correct to three decimal places, for which `g^-1(x) = g(x)`. (2 marks)
4. Calculate the area enclosed by the graphs of `g` and `g^-1`. Give your answer correct to two decimal places. (2 marks)
The diagram below shows part of the graph of the function with rule

`qquad qquad qquad f (x) = k log_e (x + a) + c`, where `k`, `a` and `c` are real constants.
- The graph has a vertical asymptote with equation `x = –1`.
- The graph has a y-axis intercept at 1.
- The point `P` on the graph has coordinates `(p, 10)`, where `p` is another real constant.

1. State the value of `a`. (1 mark)
2. Find the value of `c`. (1 mark)
3. Show that `k = 9/(log_e (p + 1)`. (2 marks)
4. Show that the gradient of the tangent to the graph of `f` at the point `P` is `9/((p + 1) log_e (p + 1))`. (1 mark)
5. If the point `(– 1, 0)` lies on the tangent referred to in part b.iv., find the exact value of `p`. (2 marks)

Show Answers Only

i. `g^-1(x) = e^((x-1)/2)-4,\ \ x in R`
ii.
iii. `– 3.914 or 5.503`
iv. `52.63\ text(units²)`
i. `1`
ii. `1`
iii. `text(Proof)\ \ text{(See Worked Solutions)}`
iv. `text(Proof)\ \ text{(See Worked Solutions)}`
v. `e^(9/10)-1`

Show Worked Solution

a.i. `text(Let)\ \ y = g(x)`

`text(Inverse: swap)\ \ x harr y,\ \ text(Domain)\ (g^-1) = text(Range)\ (g)`

`x = 2 log_e (y + 4) + 1`

`:. g^{-1} (x) = e^((x-1)/2)-4,\ \ x in R`

ii.

iii. `text(Intercepts of a function and its inverse occur)`

`text(on the line)\ \ y=x.`

`text(Solve:)\ \ g(x) = g^{-1} (x)\ \ text(for)\ \ x`

`:. x dot = – 3.914 or x = 5.503\ \ text{(3 d.p.)}`

iv.	`text(Area)`	`= int_(-3.91432…)^(5.50327…) (g(x)-g^-1 (x))\ dx`
		`= 52.63\ text{u² (2 d.p.)}`

b.i. `text(Vertical Asymptote:)`

`x = – 1`

`:. a = 1`

ii. `text(Solve)\ \ f(0) = 1\ \ text(for)\ \ c,`

`c = 1`

iii. `f(x)= k log_e (x + 1) + 1`

`text(S)text(ince)\ \ f(p)=10,`

`k log_e (p + 1) + 1`	`= 10`
`k log_e (p + 1)`	`= 9`
`:. k`	`= 9/(log_e (p + 1))\ text(… as required)`

iv.	`f^{′}(x)`	`= k/(x + 1)`
	`f^{′}(p)`	`= k/(p + 1)`
		`= (9/(log_e(p + 1))) xx 1/(p + 1)\ \ \ text{(using part (iii))}`
		`= 9/((p + 1)log_e(p + 1))\ text(… as required)`

v. `text(Two points on tangent line:)`

♦♦ Mean mark 33%.
MARKER’S COMMENT: Many students worked out the equation of the tangent which was unnecessary and time consuming.

`(p, 10),\ \ (– 1, 0)`

`f^{′} (p)`	`= (10-0)/(p-(– 1))`
	`=10/(p+1)`

`text(Solve:)\ \ 9/((p + 1)log_e(p + 1))=10/(p+1)\ \ \ text(for)\ p,`

`:.p= e^(9/10)-1`