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Popular >

extreme f(x)=12x^2-48x+32

Solution

extreme points

f (x) = 12 x^{2} - 48 x + 32

Solution

Minimum (2, - 16)

Solution steps

f^{'} (x) = 24 x - 48

Find intervals:

Decreasing

: - \infty < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 2

into

12 x^{2} - 48 x + 32 : - 16

Minimum (2, - 16)

Popular Examples

extreme f(x)= 12/7 x^7-60x^5+576x^3 extreme points

f (x) = \frac{12}{7} x^{7} - 60 x^{5} + 576 x^{3}

extreme f(x)= 1/x-ln(x)extreme points

f (x) = \frac{1}{x} - ln (x)

extreme f(x,y)=e^{5x^2+5y^2+3}extreme points

f (x, y) = e^{5 x^{2} + 5 y^{2} + 3}

extreme f(x)=-2x^3+42x^2-240x+5 extreme points

f (x) = - 2 x^{3} + 42 x^{2} - 240 x + 5

extreme y=6x-x^2-2 extreme points

y = 6 x - x^{2} - 2

Frequently Asked Questions (FAQ)

What is the extreme f(x)=12x^2-48x+32 ?
The extreme f(x)=12x^2-48x+32 is Minimum(2,-16)