What is the extreme f(x)=2x^3+45x^2-300,4<= x<= 11 ?

The extreme f(x)=2x^3+45x^2-300,4<= x<= 11 is Minimum(4,548),Maximum(11,7807)

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extreme f(x)=2x^3+45x^2-300,4<= x<= 11

extreme points

f (x) = 2 x^{3} + 45 x^{2} - 300, 4 \leq x \leq 11

Minimum (4, 548), Maximum (11, 7807)

Solution steps

Find the critical points:

x = 4, x = 11

Domain of

2 x^{3} + 45 x^{2} - 300, 4 \leq x \leq 11 : 4 \leq x \leq 11

Find intervals:

Increasing

: 4 < x < 11

Plug the extreme point

x = 4

into

2 x^{3} + 45 x^{2} - 300 \Rightarrow y = 548

Minimum (4, 548)

Plug the extreme point

x = 11

into

2 x^{3} + 45 x^{2} - 300 \Rightarrow y = 7807

Maximum (11, 7807)

Minimum (4, 548), Maximum (11, 7807)

lo g_{10} (s) (r)

f (x) = x - ln (9 x)

8 x + (\frac{1}{2 x - 1})

\frac{x ^{2} + 5 x + 11}{x - 2}

f (x, y) = 2 x^{4} - x^{2} + 3 y^{2}

What is the extreme f(x)=2x^3+45x^2-300,4<= x<= 11 ?
The extreme f(x)=2x^3+45x^2-300,4<= x<= 11 is Minimum(4,548),Maximum(11,7807)