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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i. 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)ii. 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. \(\text{As }x\to -\infty,\ \ 2^x\to 0\)
\(\therefore\ 2^x+5\to 5\)
| b. | \(g(x)\) | \(=2^x+5\) |
| \(=\Big(e^{\log_{e}{2}}\Big)^x\) | ||
| \(=e\ ^{x\log_{e}{2}}+5\) | ||
| \(g^{\prime}(x)\) | \(=\log_{e}{2}\times e\ ^{x\log_{e}{2}}\) | |
| \(=\log_{e}{2}\times 2^x\) | ||
| \(\therefore\ k\) | \(=\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\) |
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. | \(h(x)\) | \(=2^x-x^2\) |
| \(h^{\prime}(x)\) | \(=\log_{e}{(2)}\cdot 2^x-2x\ \ \text{(Using CAS)}\) | |
| \(h^{”}(x)\) | \(=(\log_{e}{(2)})^2\cdot 2^x-2\ \ \text{(Using CAS)}\) |
\(\text{Solving }h^{”}(x)=0\ \text{using CAS }\rightarrow\ x\approx 2.05753\dots\)
\(\text{Substituting into }h(x)\ \rightarrow\ h(2.05753\dots)\approx-0.070703\dots\)
\(\therefore\ \text{Point of inflection at }(2.06 , -0.07)\ \text{ correct to 2 decimal places.}\)
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. \(\text{Newton’s Method }\Rightarrow\ x_a-\dfrac{h(x_a)}{h'(x_a)}\)
\(\text{for }a=0, 1, 2, 3\ \text{given an initial estimation for }x_0=0\)
\(h(x)=2x-x^2\ \text{and }h^{\prime}(x)=\ln{2}\times 2^x-2x\)
\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 \\
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\end{array}
g. \(\text{The denominator in Newton’s Method is}\ h^{\prime}(x)=\log_{e}{(2)}\cdot 2^x-2x\)
\(\text{and the calculation will be undefined if }h^{\prime}(x)=0\ \text{as the tangent lines are horizontal}.\)
\(\therefore\ \text{The solution to }h^{\prime}(x)=0\ \text{cannot be used for }x_0.\)
h. \(\text{For a local minimum }f(x)=0\)
\(\rightarrow\ n^x-x^n=0\)
\(\rightarrow\ n^x=x^n\ \ \ (1)\)
\(\text{Also for a local minimum }f^{\prime}(x)=0\)
\(\rightarrow\ \ln(n)\cdot n^x-nx^{n-1}=0\ \ \ (2)\)
\(\text{Substitute (1) into (2)}\)
\(\ln(n)\cdot x^n-nx^{n-1}=0\)
\(x^n\Big(\ln(n)-\dfrac{n}{x}\Big)=0\)
\(\therefore\ x^n=0\ \text{or }\ \ln(n)=\dfrac{n}{x}\)
\(x=0\ \text{or }x=\dfrac{n}{\ln(n)}\)
\(\therefore\ n=e\)
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.