# (Application of the derivative)

Critical Points

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**Definition of critical points or critical numbers :**

Let 'f' be defined at c. If f '(c) = 0 or if is not differentiable at 'c' then 'c' is a critical number of f.

**Steps to find the critical numbers :**

Let f(x) be the given function

1) Find f '(x) or the 1st derivative of the given function.

2) Set f'(x) equal to zero.

3) Find the value of x ,that gives us the critical numbers 'c' such that f '(c) = 0.

## Examples on Critical Points

**Example 1:**Find the critical numbers of f(x) = $x^{3} -6x^{2} + 9x - 8 $

**Solution :**f(x) = $x^{3} -6x^{2} + 9x - 8 $

Find the derivative with respect to x

f'(x) = $3x^{2} -12x + 9 $

Set f'(x) = 0

$3x^{2} -12x + 9 $ = 0

Since this is a quadratic equation so we will factor it.

$3(x^{2} -4x + 3 ) $ = 0

3 (x - 3) (x - 1) = 0

x = 1,3

Critical points c = 1,3

**Example 2:**Find the critical numbers of f(x) = $x + e^{-x} $

**Solution :**f(x) = $x + e^{-x} $

Find the derivative with respect to x

f'(x) = $1 + e^{-x}.(-1) $

f'(x) = $1 - e^{-x}$

Set f'(x) = 0

$1 - e^{-x} $ = 0

$e^{-x} $ = 1

Take ln on both sides,

$ln(e^{-x}) = ln(1)$

-x ln(e) = 0

-x .1 = 0

x = 0

Critical points c = 0

**Critical points are used to find the extrema on a closed interval [a,b].**

Apply the following steps :

1) Find the critical points of in (a,b).

2) Evaluate at each critical number in (a,b)

3) Evaluate at each endpoint of [a,b].

4) The least of these values is the local minimum and the greatest is the local maximum.

**12th grade math**

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