What is the extreme f(x)=x^{4/7}(3x+11) ?

The extreme f(x)=x^{4/7}(3x+11) is Maximum(-4/3 ,7(-4/3)^{4/7}),Minimum(0,0)

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extreme f(x)=x^{4/7}(3x+11)

extreme points

f (x) = x^{\frac{4}{7}} (3 x + 11)

Maximum (- \frac{4}{3}, 7 (- \frac{4}{3})^{\frac{4}{7}}), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{11 ( 3 x + 4 )}{7 x ^{\frac{3}{7}}}

Find intervals:

Increasing

: - \infty < x < - \frac{4}{3},

Decreasing

: - \frac{4}{3} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{4}{3}

into

x^{\frac{4}{7}} (3 x + 11) : 7 (- \frac{4}{3})^{\frac{4}{7}}

Maximum (- \frac{4}{3}, 7 (- \frac{4}{3})^{\frac{4}{7}})

Plug

x = 0

into

x^{\frac{4}{7}} (3 x + 11) : 0

Minimum (0, 0)

Maximum (- \frac{4}{3}, 7 (- \frac{4}{3})^{\frac{4}{7}}), Minimum (0, 0)

f (x) = - 0.2 x^{2} + 1.6 x + 98.6

6 x + 18

f (x) = 5 x^{2} - x - 2

f (x) = x^{3} - 6 x^{2} + 5 [- 3.5]

p (x) = \frac{x ^{4} - 2 x ^{3} + 3 x ^{2} - a x + 3 a - 7}{x + 2}

What is the extreme f(x)=x^{4/7}(3x+11) ?
The extreme f(x)=x^{4/7}(3x+11) is Maximum(-4/3 ,7(-4/3)^{4/7}),Minimum(0,0)