What is the extreme f(x)=-0.02x^2+0.39x-1.07 ?

The extreme f(x)=-0.02x^2+0.39x-1.07 is Maximum(39/4 ,0.83125)

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extreme f(x)=-0.02x^2+0.39x-1.07

extreme

Maximum (\frac{39}{4}, 0.83125)

Solution steps

f^{'} (x) = - 0.04 x + 0.39

Find intervals:

Increasing

: - \infty < x < \frac{39}{4},

Decreasing

: \frac{39}{4} < x < \infty

Plug

x = \frac{39}{4}

into

- 0.02 x^{2} + 0.39 x - 1.07 : 0.83125

Maximum (\frac{39}{4}, 0.83125)

What is the extreme f(x)=-0.02x^2+0.39x-1.07 ?
The extreme f(x)=-0.02x^2+0.39x-1.07 is Maximum(39/4 ,0.83125)