(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.

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