What is the extreme f(x)=x^3-24ln(x)+5 ?

The extreme f(x)=x^3-24ln(x)+5 is Minimum(2,-24ln(2)+13)

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extreme f(x)=x^3-24ln(x)+5

extreme points

f (x) = x^{3} - 24 ln (x) + 5

Minimum (2, - 24 ln (2) + 13)

Solution steps

f^{'} (x) = 3 x^{2} - \frac{24}{x}

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 2

into

x^{3} - 24 ln (x) + 5 : - 24 ln (2) + 13

Minimum (2, - 24 ln (2) + 13)

f (x) = 16 x + 19 x^{\frac{16}{19}}

f (t) = t^{2} + \frac{1}{t}

y = \frac{1}{3} x^{3} + 2 x^{2} + 3 x

f (x) = y \cdot (e^{x} - 1)

f (x, y) = 2 x^{2} + 8 x - 3 y^{2} + 24 y + 7

What is the extreme f(x)=x^3-24ln(x)+5 ?
The extreme f(x)=x^3-24ln(x)+5 is Minimum(2,-24ln(2)+13)