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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- 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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- 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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- 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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- Hence, or otherwise, find the equation of the tangent to \(g\) that passes through the origin, correct to three decimal places. (2 marks)
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. \(\text{As }x\to -\infty,\ \ 2^x\to 0\)
\(\therefore\ 2^x+5\to 5\)
| 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. \(\text{Tangent at}\ (a, g(a)):\)
| \(y-(2^a+5)\) | \(=\log_{e}{2}\times2^a(x-a)\) |
| \(\therefore\ y\) | \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\) |
cii. \(\text{Substitute }(0, 0)\ \text{into equation from c(i) to find}\ a\)
| \( y\) | \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\) |
| \(0\) | \(=2^a\ \log_{e}{(2)\times 0}-a\ 2^a\ \log_{e}{(2)}+2^a+5\) |
| \(0\) | \(=-a\ 2^a\ \log_{e}{(2)}+2^a+5\) |
\(\text{Solve for }a\text{ using CAS }\rightarrow\ a\approx 2.61784\dots\)
\(\text{Equation of tangent when }\ a\approx 2.6178\)
| \( y\) | \(=2^{2.6178..}\ \log_{e}{(2)x}+0\) |
| \(\therefore\ y\) | \(=4.255x\) |
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. \(\text{From graph (CAS), }h(x)\ \text{is strictly decreasing between the 2 turning points.}\)
\(\therefore\ \text{Largest interval includes endpoints and is given by }\rightarrow\ [0.49, 3.21]\)
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. \(\text{The denominator in Newton’s Method is}\ h'(x)=\log_{e}{(2)}\cdot 2^x-2x\)
\(\text{and the calculation will be undefined if }h'(x)=0\ \text{as the tangent lines are horizontal}.\)
\(\therefore\ \text{The solution to }h'(x)=0\ \text{cannot be used for }x_0.\)
h. \(\text{For a local minimum }f(x)=0\)
\(\rightarrow\ n^x-x^n=0\)
\(\rightarrow\ n^x=x^n\ \ \ (1)\)
\(\text{Also for a local minimum }f'(x)=0\)
\(\rightarrow\ \ln(n)\cdot n^x-nx^{n-1}=0\ \ \ (2)\)
\(\text{Substitute (1) into (2)}\)
\(\ln(n)\cdot x^n-nx^{n-1}=0\)
\(x^n\Big(\ln(n)-\dfrac{n}{x}\Big)=0\)
\(\therefore\ x^n=0\ \text{or }\ \ln(n)=\dfrac{n}{x}\)
\(x=0\ \text{or }x=\dfrac{n}{\ln(n)}\)
\(\therefore\ n=e\)
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.