What is the extreme f(x)=(x^3)/x-3x^2-7x ?

The extreme f(x)=(x^3)/x-3x^2-7x is Maximum(-7/4 , 49/8)

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extreme f(x)=(x^3)/x-3x^2-7x

extreme

Maximum (- \frac{7}{4}, \frac{49}{8})

Solution steps

f^{'} (x) = - 4 x - 7

Find intervals:

Increasing

: - \infty < x < - \frac{7}{4},

Decreasing

: - \frac{7}{4} < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - \frac{7}{4}

into

\frac{x ^{3}}{x} - 3 x^{2} - 7 x : \frac{49}{8}

Maximum (- \frac{7}{4}, \frac{49}{8})

What is the extreme f(x)=(x^3)/x-3x^2-7x ?
The extreme f(x)=(x^3)/x-3x^2-7x is Maximum(-7/4 , 49/8)