What is the extreme f(x)=340x^2+4080x^3,0<= x<= 0.08 ?

The extreme f(x)=340x^2+4080x^3,0<= x<= 0.08 is Minimum(0,0),Maximum(0.08,4.26496)

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extreme f(x)=340x^2+4080x^3,0<= x<= 0.08

extreme points

f (x) = 340 x^{2} + 4080 x^{3}, 0 \leq x \leq 0.08

Minimum (0, 0), Maximum (0.08, 4.26496)

Solution steps

Find the critical points:

x = 0, x = 0.08

Domain of

340 x^{2} + 4080 x^{3}, 0 \leq x \leq 0.08 : 0 \leq x \leq 0.08

Find intervals:

Increasing

: 0 < x < 0.08

Plug the extreme point

x = 0

into

340 x^{2} + 4080 x^{3} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 0.08

into

340 x^{2} + 4080 x^{3} \Rightarrow y = 4.26496

Maximum (0.08, 4.26496)

Minimum (0, 0), Maximum (0.08, 4.26496)

y = - x^{2} - 6 x - 8

f (x) = x e^{- \frac{x ^{2}}{162}}

f (x, y) = x^{2} + 2 \cdot y^{2} + 4 \cdot x - 4 \cdot y

f (x) = x^{4} - 8 x^{2} + 18

f (x) = x^{3} + 29 x + 128

What is the extreme f(x)=340x^2+4080x^3,0<= x<= 0.08 ?
The extreme f(x)=340x^2+4080x^3,0<= x<= 0.08 is Minimum(0,0),Maximum(0.08,4.26496)