The graph intersects the x-axis at \(x=1\) and has a stationary point at \(x=-1\).
(i) Find the x-coordinate of the other stationary point.
(ii) Determine what type of stationary point this is.
Try the question yourself first. If you get stuck, open the hints before using the full walkthrough.
Differentiate first, use \(f'(-1)=0\) to link \(a\) and \(b\), then use \(f(1)=0\) to solve for the constants before classifying the other stationary point with the second derivative.
Step 1
Differentiate the function
What is \(f'(x)\)?
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Step 2
Use the stationary point at \(x=-1\)
Since \(x=-1\) is a stationary point, \(f'(-1)=0\). What relation does that give?
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Step 3
Use the x-intercept at \(x=1\)
Since the graph crosses the x-axis at \(x=1\), we know \(f(1)=0\). Which equation does that give?
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Step 4
Solve for \(a\) and \(b\)
Substitute \(a=\frac{b}{2}\) into \(2a+b+2=0\). Enter your answer as \((a,b)\).
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Step 5
Find the other stationary point
Substitute the values of \(a\) and \(b\) into \(f'(x)\) and solve \(f'(x)=0\).
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Step 6
Use the second derivative test
What is the nature of the stationary point at \(x=-\frac{1}{3}\)?
Step 7
Final answers
Now write both parts clearly.
\[
f'(x)=3ax^2+2bx+a
\]
\[
f'(-1)=0 \Rightarrow 3a-2b+a=0
\]
\[
4a-2b=0 \Rightarrow a=\frac{b}{2}
\]
\[
f(1)=0 \Rightarrow a+b+a+2=0
\]
\[
2a+b+2=0
\]
\[
2\left(\frac{b}{2}\right)+b+2=0
\]
\[
2b+2=0 \Rightarrow b=-1,\quad a=-\frac{1}{2}
\]
\[
f'(x)=-\frac{3}{2}x^2-2x-\frac{1}{2}
\]
\[
3x^2+4x+1=0
\]
\[
(3x+1)(x+1)=0
\]
\[
x=-1 \text{ or } x=-\frac{1}{3}
\]
\[
\text{So the other stationary point is at }x=-\frac{1}{3}
\]
\[
f''(x)=6ax+2b=-3x-2
\]
\[
f''\left(-\frac{1}{3}\right)=1-2=-1<0
\]
\[
\text{So it is a local maximum.}
\]
Great work. The other stationary point is at \(x=-\frac{1}{3}\), and it is a local maximum.