What is the extreme f(x)=4x^3+21x^2-294x+9 ?

The extreme f(x)=4x^3+21x^2-294x+9 is Maximum(-7,1724),Minimum(7/2 ,-2365/4)

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extreme f(x)=4x^3+21x^2-294x+9

extreme points

f (x) = 4 x^{3} + 21 x^{2} - 294 x + 9

Maximum (- 7, 1724), Minimum (\frac{7}{2}, - \frac{2365}{4})

Solution steps

f^{'} (x) = 12 x^{2} + 42 x - 294

Find intervals:

Increasing

: - \infty < x < - 7,

Decreasing

: - 7 < x < \frac{7}{2},

Increasing

: \frac{7}{2} < x < \infty

Plug

x = - 7

into

4 x^{3} + 21 x^{2} - 294 x + 9 : 1724

Maximum (- 7, 1724)

Plug

x = \frac{7}{2}

into

4 x^{3} + 21 x^{2} - 294 x + 9 : - \frac{2365}{4}

Minimum (\frac{7}{2}, - \frac{2365}{4})

Maximum (- 7, 1724), Minimum (\frac{7}{2}, - \frac{2365}{4})

What is the extreme f(x)=4x^3+21x^2-294x+9 ?
The extreme f(x)=4x^3+21x^2-294x+9 is Maximum(-7,1724),Minimum(7/2 ,-2365/4)