# Consider the function f(x) = x^2 - 6x a. Find the critical numbers of the function. b. Find...

## Question:

Consider the function

{eq}f(x) = x^2 - 6x {/eq}

a. Find the critical numbers of the function.

b. Find the open intervals where the function is increasing or decreasing.

c. Apply the First Derivative Test to identify the relative extremum.

## Maxima and minima:

Maxima and minima of a function are used to find the maximum and minimum value of the function. In critical points, the function either increasing to decreasing or decreasing to increasing. In this problem, we will find using the first derivative test.

## Answer and Explanation:

Given that {eq}f(x) = x^2 - 6x {/eq}

(a) Differentiate with respect to 'x' we get

{eq}f'\left(x\right)=2x-6\\ f\:'\left(x\right)=0\\ 2x-6=0\\ 2x=6\\ x=3 {/eq}

The critical number is {eq}x=3 {/eq}

(b) The given interval divided into {eq}-\infty \:<x<3 {/eq} and {eq}3<x<\infty {/eq}

(c) Apply first derivative test, we get {eq}\mathrm{The\:sign\:of\:} f\:'\left(x\right) \mathrm{\:at\:}-\infty \:<x<3 \mathrm{\:is \: Negative}\\ \Rightarrow f(x) \, is \, decreasing\\ \mathrm{The\:sign\:of\:} f\:'\left(x\right) \mathrm{\:at\:}3<x<\infty \mathrm{\: is \: Positive}\\ \Rightarrow f(x) \, is \, increasing\\ {/eq}

The Minimum value occur at {eq}(3, -9) {/eq}

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#### Learn more about this topic:

from Math 104: Calculus

Chapter 9 / Lesson 3