What is the extreme f(x)= 1/20 x^2-25sqrt(x) ?

The extreme f(x)= 1/20 x^2-25sqrt(x) is Maximum(0,0),Minimum(25,-375/4)

Popular >

extreme f(x)= 1/20 x^2-25sqrt(x)

extreme points

f (x) = \frac{1}{20} x^{2} - 25 x

Maximum (0, 0), Minimum (25, - \frac{375}{4})

Solution steps

Find the critical points:

x = 0, x = 25

Domain of

\frac{1}{20} x^{2} - 25 x : x \geq 0

Find intervals:

Decreasing

: 0 < x < 25,

Increasing

: 25 < x < \infty

Plug the extreme point

x = 0

into

\frac{1}{20} x^{2} - 25 x \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 25

into

\frac{1}{20} x^{2} - 25 x \Rightarrow y = - \frac{375}{4}

Minimum (25, - \frac{375}{4})

Maximum (0, 0), Minimum (25, - \frac{375}{4})

f (x) = 3 x^{3} + \frac{9}{2} x^{2} - 54 x + 1

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

f (x) = \frac{x ^{4} - 5 x + 7}{x ^{2} + 1}

f (x) = 5 - 6 x^{2}

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

What is the extreme f(x)= 1/20 x^2-25sqrt(x) ?
The extreme f(x)= 1/20 x^2-25sqrt(x) is Maximum(0,0),Minimum(25,-375/4)