What is the extreme f(x)=-16x^2+60x+2 ?

The extreme f(x)=-16x^2+60x+2 is Maximum(15/8 , 233/4)

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extreme f(x)=-16x^2+60x+2

extreme

Maximum (\frac{15}{8}, \frac{233}{4})

Solution steps

f^{'} (x) = - 32 x + 60

Find intervals:

Increasing

: - \infty < x < \frac{15}{8},

Decreasing

: \frac{15}{8} < x < \infty

Plug

x = \frac{15}{8}

into

- 16 x^{2} + 60 x + 2 : \frac{233}{4}

Maximum (\frac{15}{8}, \frac{233}{4})

What is the extreme f(x)=-16x^2+60x+2 ?
The extreme f(x)=-16x^2+60x+2 is Maximum(15/8 , 233/4)