Prove the Second Derivative Test for minimums.

Primary tabs

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 $f\,''(e)\gt 0$, then $e$ is a minimum of $f(x)$.

Answer: 

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

SAMPLE SOLUTION

All of our problems have fully worked out, logical solutions. We connect every solution to its underlying lesson so you can understand the concepts needed to solve any problem and reinforce your learning by practicing with similar problems. All of the formulas are available to our subscribers, but here is a sample of the problem solution you will see when you log in:



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 minimum.