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Popular >

extreme f(x)=x^2-15x+56

Solution

extreme points

f (x) = x^{2} - 15 x + 56

Solution

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

Solution steps

f^{'} (x) = 2 x - 15

Find intervals:

Decreasing

: - \infty < x < \frac{15}{2},

Increasing

: \frac{15}{2} < x < \infty

Plug

x = \frac{15}{2}

into

x^{2} - 15 x + 56 : - \frac{1}{4}

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

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Frequently Asked Questions (FAQ)

What is the extreme f(x)=x^2-15x+56 ?
The extreme f(x)=x^2-15x+56 is Minimum(15/2 ,-1/4)