What is the extreme f(x)=49x^2-10x+17 ?

The extreme f(x)=49x^2-10x+17 is Minimum(5/49 , 808/49)

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extreme f(x)=49x^2-10x+17

extreme

Minimum (\frac{5}{49}, \frac{808}{49})

Solution steps

f^{'} (x) = 98 x - 10

Find intervals:

Decreasing

: - \infty < x < \frac{5}{49},

Increasing

: \frac{5}{49} < x < \infty

Plug

x = \frac{5}{49}

into

49 x^{2} - 10 x + 17 : \frac{808}{49}

Minimum (\frac{5}{49}, \frac{808}{49})

What is the extreme f(x)=49x^2-10x+17 ?
The extreme f(x)=49x^2-10x+17 is Minimum(5/49 , 808/49)