What is the extreme f(x)=(7+x)(11-3x)^{1/3} ?

The extreme f(x)=(7+x)(11-3x)^{1/3} is Maximum(1,16)

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extreme f(x)=(7+x)(11-3x)^{1/3}

extreme points

f (x) = (7 + x) (11 - 3 x)^{\frac{1}{3}}

Maximum (1, 16)

Solution steps

f^{'} (x) = - \frac{4 ( x - 1 )}{( 11 - 3 x ) ^{\frac{2}{3}}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < \frac{11}{3},

Decreasing

: \frac{11}{3} < x < \infty

Plug

x = 1

into

(7 + x) (11 - 3 x)^{\frac{1}{3}} : 16

Maximum (1, 16)

y = x^{- 4} \cdot e^{x^{4}}

f (x) = - 2 x^{3} + 12

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

f (x) = \frac{2 + 4 x ^{2}}{1 - 3 x}

f (x) = 6 x + \frac{24}{x ^{2}} + 5, x > 0

What is the extreme f(x)=(7+x)(11-3x)^{1/3} ?
The extreme f(x)=(7+x)(11-3x)^{1/3} is Maximum(1,16)