What is the extreme y=(x^4}{12}-\frac{x^3)/2+x^2+10 ?

The extreme y=(x^4}{12}-\frac{x^3)/2+x^2+10 is Minimum(0,10)

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extreme y=(x^4}{12}-\frac{x^3)/2+x^2+10

extreme points

y = \frac{x ^{4}}{12} - \frac{x ^{3}}{2} + x^{2} + 10

Minimum (0, 10)

Solution steps

f^{'} (x) = \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + 2 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

\frac{x ^{4}}{12} - \frac{x ^{3}}{2} + x^{2} + 10 : 10

Minimum (0, 10)

f (x) = (x - 2)^{7} (x + 4)^{7}

sin^{8} (x) + cos^{8} (x)

f (x, y) = 400 - 81 x^{2} - 4 y^{2}

f (x) = 9 t^{2} + 4 t - 10

(x^{2} - 4)^{\frac{1}{7}}, - 2 \leq x \leq 3

What is the extreme y=(x^4}{12}-\frac{x^3)/2+x^2+10 ?
The extreme y=(x^4}{12}-\frac{x^3)/2+x^2+10 is Minimum(0,10)