# Find the local max and min values of f using the first and second derivative tests. f(x)=1 +...

## Question:

Find the local max and min values of f using the first and second derivative tests.

{eq}f(x)=1 + 3x^2 - 2x^2 {/eq}

## Local and Global Extreme:

With the second derivative test we can sometimes confirm whether we have a maximum or a minimum at the critical point:

if this sign is constant, we can assure that this local extreme is also global.

## Answer and Explanation:

First, with the first derivative, we can calculate the critical points:

{eq}f\left( x \right) = 1 + 3{x^2} - 2{x^2} = 1 + {x^2}\\ \\ f'\left( x \right) = 2x\\ 2x = 0 \to x = \frac{0}{{2}} = 0 {/eq}

As the second derivative is negative, a local (and at this case global) minimum is reached:

{eq}f''\left( x \right) = 2 > 0 \to \min {/eq}

That is, the minimum values is 1: {eq}f\left( 0 \right) = 1 + {0^2} = 1. {/eq}