What is the extreme f(x)=315x^2+1260x^3,0<= x<= 0.25 ?

The extreme f(x)=315x^2+1260x^3,0<= x<= 0.25 is Minimum(0,0),Maximum(0.25,39.375)

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extreme f(x)=315x^2+1260x^3,0<= x<= 0.25

extreme points

f (x) = 315 x^{2} + 1260 x^{3}, 0 \leq x \leq 0.25

Minimum (0, 0), Maximum (0.25, 39.375)

Solution steps

Find the critical points:

x = 0, x = 0.25

Domain of

315 x^{2} + 1260 x^{3}, 0 \leq x \leq 0.25 : 0 \leq x \leq 0.25

Find intervals:

Increasing

: 0 < x < 0.25

Plug the extreme point

x = 0

into

315 x^{2} + 1260 x^{3} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 0.25

into

315 x^{2} + 1260 x^{3} \Rightarrow y = 39.375

Maximum (0.25, 39.375)

Minimum (0, 0), Maximum (0.25, 39.375)

f (x, y) = 34 - 9 x^{2} - 4 y^{2}

f (x, y) = ln (y^{2} + x) - \frac{1}{2} x + y

(x - 5) e^{x}

x^{\frac{6}{7}} - 7

What is the extreme f(x)=315x^2+1260x^3,0<= x<= 0.25 ?
The extreme f(x)=315x^2+1260x^3,0<= x<= 0.25 is Minimum(0,0),Maximum(0.25,39.375)