What is the extreme f(x)=1x-4/3 x^{3/4},(0,infinity) ?

The extreme f(x)=1x-4/3 x^{3/4},(0,infinity) is Minimum(1,-1/3)

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extreme f(x)=1x-4/3 x^{3/4},(0,infinity)

extreme points

f (x) = 1 x - \frac{4}{3} x^{\frac{3}{4}}, (0, \infty)

Minimum (1, - \frac{1}{3})

Solution steps

Find the critical points:

x = 1

Domain of

1 x - \frac{4}{3} x^{\frac{3}{4}}, (0, \infty) : x > 0

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug the extreme point

x = 1

into

1 x - \frac{4}{3} x^{\frac{3}{4}} \Rightarrow y = - \frac{1}{3}

Minimum (1, - \frac{1}{3})

f (x) = \frac{25 x}{x + 2} - x

y = x^{2} - 12

f (x) = 12 x^{2} - 180 x + 464

f (x) = - \frac{3}{16} x^{\frac{8}{3}} + \frac{3}{4} x^{\frac{2}{3}}

2 sin (x) - cos (x) - 4 x

What is the extreme f(x)=1x-4/3 x^{3/4},(0,infinity) ?
The extreme f(x)=1x-4/3 x^{3/4},(0,infinity) is Minimum(1,-1/3)