Let

1. Show that

   $$f^{\prime }(x)=f(x)\frac{2x}{2x+1}$$

2. Hence, show that

   $$f^{″}(x)=2f(x)\frac{2x^2+1}{(2x+1{)}^2}$$

3. Hence, find the coordinates of, and determine the nature of, the turning point of the curve

   $$y=\frac{e^x}{\sqrt{2x+1}}$$

For (a), use the quotient rule, and then try to spot $f(x)$ within your answer.

When doing part (b), don't replace $f(x)$ with $\frac{e^x}{\sqrt{2x+1}}$. Rather, leave it as $f(x)$ and use the product rule so that you get

You know $f^{\prime }(x)$ from part (a)!

Let the turning point have coordinates $(p,q)$.

If it is a minimum, give $17(p+1)(q+1)$. If it is a maximum, give $19(p+1)(q+1)$.