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Popular >

extreme f(x)=e^{x^2-9}

Solution

extreme points

f (x) = e^{x^{2} - 9}

Solution

Minimum (0, \frac{1}{e ^{9}})

Solution steps

f^{'} (x) = e^{x^{2} - 9} \cdot 2 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

e^{x^{2} - 9} : \frac{1}{e ^{9}}

Minimum (0, \frac{1}{e ^{9}})

Popular Examples

extreme f(x)=5ye^x-6e^y extreme points

f (x) = 5 y e^{x} - 6 e^{y}

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f (x) = x_{2} - 4 \cdot 2 - 4 x

extreme f(x)=x^2-9ln(x)extreme points

f (x) = x^{2} - 9 ln (x)

extreme f(x)=(1/12 x^3+x^2+3x+26/3)extreme points

f (x) = (\frac{1}{12} x^{3} + x^{2} + 3 x + \frac{26}{3})

extreme f(x)=-5x^3+45x extreme points

f (x) = - 5 x^{3} + 45 x

Frequently Asked Questions (FAQ)

What is the extreme f(x)=e^{x^2-9} ?
The extreme f(x)=e^{x^2-9} is Minimum(0, 1/(e^9))