The graph below shows \(y=f(x)\). Use it to sketch a possible graph of \(y=f'(x)\).
Sketch \(f'(x)\) here:
Try the question yourself first. If you get stuck, open the hints before using the full walkthrough.
The key idea is that stationary points of \(f\) become x-intercepts of \(f'\). Then use where \(f\) is increasing, decreasing, and changing concavity to shape the derivative.
Step 1
Find the stationary points of \(f\)
Look for the places where the tangent to \(f\) is horizontal.
Step 2
Turn those into x-intercepts of \(f'(x)\)
If \(f\) has a horizontal tangent, then \(f'(x)=0\) there. Where should the derivative cross the x-axis?
Step 3
Work out the sign of \(f'(x)\)
Use whether \(f\) is increasing or decreasing on each interval.
Step 4
Place the turning points of \(f'(x)\)
The turning points of the derivative happen where \(f\) changes concavity.
Step 5
Decide the end behaviour
How should the ends of the derivative graph look?
Step 6
Final sketch
A good sketch of \(f'(x)\) is a cubic-like curve crossing the x-axis at the stationary points of \(f\), with signs matching the increasing/decreasing intervals.
\[
f'(x)=0 \text{ at about } x=-3,\ 0,\text{ and }4
\]
\[
f'(x)<0 \text{ before the first turning point of }f
\]
\[
f'(x)>0 \text{ between the first minimum and the maximum of }f
\]
\[
f'(x)<0 \text{ between the maximum and the second minimum of }f
\]
\[
f'(x)>0 \text{ after the second minimum of }f
\]
The exact heights of the derivative graph are not fixed by the original sketch. What matters most is the correct x-intercepts, sign changes, and overall cubic shape.
One possible sketch of \(f'(x)\)
Turn the markers on and off to see how the stationary points of \(f\) determine the x-intercepts of \(f'\), and how changes in concavity place the turning points of the derivative.