Science:Math Exam Resources/Courses/MATH110/April 2019/Question 01 (i)/Solution 1

Since $f$ is differentiable everywhere, $f'(1)=0$. Note also that $f(1)=0$ since $(1,0)$ is a point. The derivative of $g(x)$ is

$g'(x)=xf'(x)+f(x)+1.$

Evaluated at $x=1$ we have

$g'(1)=f'(1)+f(1)+1=0+0+1=1.$ Here, we know that $f(1)=0$ because we are told that $(1,f(1))=(1,0)$, and we also know that $f$ has a local maximum at $(1,0)$, which means that $f^{'}(1)=0$.

Since the derivative $g'(1)$ is not zero, $g$ does not have a local extreme at $x=1$.