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Popular >

extreme f(x)=x^4-32x+5

Solution

extreme points

f (x) = x^{4} - 32 x + 5

Solution

Minimum (2, - 43)

Solution steps

f^{'} (x) = 4 x^{3} - 32

Find intervals:

Decreasing

: - \infty < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 2

into

x^{4} - 32 x + 5 : - 43

Minimum (2, - 43)

Popular Examples

parity f(x)=2+tan(x)parity

f (x) = 2 + tan (x)

parallel y-(-8)=5(x-3)parallel

y - (- 8) = 5 (x - 3)

range of f(x)= 1/(sqrt(x+5))range

f (x) = \frac{1}{x + 5}

critical f(x)=(480)/(x^6)critical points

f (x) = \frac{480}{x ^{6}}

extreme f(x)= 5/(x-6)extreme points

f (x) = \frac{5}{x - 6}

Frequently Asked Questions (FAQ)

What is the extreme f(x)=x^4-32x+5 ?
The extreme f(x)=x^4-32x+5 is Minimum(2,-43)