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Popular >

extreme y=2x^2-8x+8

Solution

extreme points

y = 2 x^{2} - 8 x + 8

Solution

Minimum (2, 0)

Solution steps

f^{'} (x) = 4 x - 8

Find intervals:

Decreasing

: - \infty < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 2

into

2 x^{2} - 8 x + 8 : 0

Minimum (2, 0)

Graph

Popular Examples

extreme f(x,y)=(4x^2+10y^2)e^{-y}extreme points

f (x, y) = (4 x^{2} + 10 y^{2}) e^{- y}

extreme P(x,y)=6x^2-xy-y^2+y+8x+2 extreme points

P (x, y) = 6 x^{2} - x y - y^{2} + y + 8 x + 2

extreme f(x)=(x+2)^{4/3}extreme points

f (x) = (x + 2)^{\frac{4}{3}}

extreme f(x)=x^3-3x^2+1,-1/2 <= x<= 4 extreme points

f (x) = x^{3} - 3 x^{2} + 1, - \frac{1}{2} \leq x \leq 4

extreme f(x)=(x^2-10x+25)/(x-10)extreme points

f (x) = \frac{x ^{2} - 10 x + 25}{x - 10}

Frequently Asked Questions (FAQ)

What is the extreme y=2x^2-8x+8 ?
The extreme y=2x^2-8x+8 is Minimum(2,0)