Question
\[
y=9x-2+\frac{3}{3x-1}
\]
Find the \(x\)-value(s) of any stationary point(s) on the graph and determine their nature.
You must use calculus and show any derivatives that you need when solving this problem.
Try the question yourself first. If you get stuck, open the hints before using the full walkthrough.
Step 1
Differentiate the function
Differentiate each term, using the chain rule on the fraction term.
Step 2
Find where the derivative is zero
Stationary points happen when \(y'=0\).
\[
0=9-\frac{9}{(3x-1)^2}
\]
\[
1=\frac{1}{(3x-1)^2}
\]
\[
(3x-1)^2=1
\]
Step 3
Use the second derivative
Differentiate again so we can classify each stationary point.
Step 4
Determine the nature
Substitute the \(x\)-values into \(y''\).
Step 5
Final answer
\[
y'=9-\frac{9}{(3x-1)^2}
\]
\[
x=0 \text{ and } x=\frac{2}{3}
\]
\[
y(0)=9(0)-2+\frac{3}{-1}=-5
\]
\[
y\left(\frac{2}{3}\right)=9\left(\frac{2}{3}\right)-2+\frac{3}{1}=7
\]
\[
y''=\frac{54}{(3x-1)^3}
\]
\[
y''(0)=-54<0 \Rightarrow \text{local maximum}
\]
\[
y''\left(\frac{2}{3}\right)=54>0 \Rightarrow \text{local minimum}
\]
So the graph has a local maximum at \((0,-5)\) and a local minimum at \(\left(\frac{2}{3},7\right)\).