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Popular >

extreme f(x)=x^2+x+4

Solution

extreme points

f (x) = x^{2} + x + 4

Solution

Minimum (- \frac{1}{2}, \frac{15}{4})

Solution steps

f^{'} (x) = 2 x + 1

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{2},

Increasing

: - \frac{1}{2} < x < \infty

Plug

x = - \frac{1}{2}

into

x^{2} + x + 4 : \frac{15}{4}

Minimum (- \frac{1}{2}, \frac{15}{4})

Graph

Popular Examples

extreme f(x)=x^2+x+1 extreme points

f (x) = x^{2} + x + 1

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

y = - 2 x^{2} + 6

extreme f(x)=7x^{2/5}+2x^{7/5}extreme points

f (x) = 7 x^{\frac{2}{5}} + 2 x^{\frac{7}{5}}

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

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

extreme f(x)=((4860)/x)+12x+709407 extreme points

f (x) = (\frac{4860}{x}) + 12 x + 709407

Frequently Asked Questions (FAQ)

What is the extreme f(x)=x^2+x+4 ?
The extreme f(x)=x^2+x+4 is Minimum(-1/2 , 15/4)