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

MATH110 April 2019

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Question 01 (i)
Suppose $f$ is differentiable everywhere and assume it has a local maximum at the point $(1,0)$ . Does the function $g(x)=xf(x)+x+3$ have a local extreme value at $x=1$ ? Justify your answer.

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Hint
The new function is differentiable at $x=1$ also. Therefore if it has a local max, then its derivative must be zero.

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $x=1$ .