Prove the Second Derivative Test for maximums.

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Lesson Parent: 
Geometry of Functions VIII: The Second Derivative Test
Problem: 

Assume that $e$ is a critical point of $f(x)$ and that $f\,'(e)=0$. Prove that if $f\,''(e)\lt 0$, then $e$ is a maximum of $f(x)$.

Answer: 

It is true that if $f\,''(e)\lt 0$, then $e$ is a maximum of $f(x)$.

SAMPLE SOLUTION

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Solution: 

Since <Sign in to see all the formulas> is the derivative of <Sign in to see all the formulas>, we know that if <Sign in to see all the formulas>, then <Sign in to see all the formulas> for all positive $k$ sufficiently small. (Think of <Sign in to see all the formulas> as a new function $g(x)$. Now if <Sign in to see all the formulas>, then <Sign in to see all the formulas> for all positive $k$ sufficiently small. Now substitute back in the fact that <Sign in to see all the formulas>.) The proof let's us know what sufficiently small means. It dictates that there is a <Sign in to see all the formulas> such that $k$ sufficiently small means <Sign in to see all the formulas>.

Since the problem assumes that <Sign in to see all the formulas>, <Sign in to see all the formulas> and <Sign in to see all the formulas>. And since <Sign in to see all the formulas>, there is an interval where <Sign in to see all the formulas> for all $x$ in <Sign in to see all the formulas> and <Sign in to see all the formulas> for all $x$ in <Sign in to see all the formulas>. Therefore, $f(x)$ in the interval <Sign in to see all the formulas> satisfies the precise requirement for the First Derivative Test and therefore $f(e)$ is a maximum.