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Popular >

extreme f(1,y)=y^2+y+6

Solution

extreme points

f (1, y) = y^{2} + y + 6

Solution

Minimum (- \frac{1}{2}, \frac{23}{4})

Solution steps

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

Find intervals:

Decreasing

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

Increasing

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

Plug

y = - \frac{1}{2}

into

y^{2} + y + 6 : \frac{23}{4}

Minimum (- \frac{1}{2}, \frac{23}{4})

Popular Examples

extreme f(x)=-10x^2+60x-80 extreme points

f (x) = - 10 x^{2} + 60 x - 80

extreme f(x)=-5/(x-2)extreme points

f (x) = - \frac{5}{x - 2}

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

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

extreme f(x)=(x^2-9)^{1/3}extreme points

f (x) = (x^{2} - 9)^{\frac{1}{3}}

extreme f(x)=x^4-8x^3+20 extreme points

f (x) = x^{4} - 8 x^{3} + 20

Frequently Asked Questions (FAQ)

What is the extreme f(1,y)=y^2+y+6 ?
The extreme f(1,y)=y^2+y+6 is Minimum(-1/2 , 23/4)