What is the extreme f(x)=-3x^2+180x+190,0<= x<= 40 ?

The extreme f(x)=-3x^2+180x+190,0<= x<= 40 is Minimum(0,190),Maximum(30,2890),Minimum(40,2590)

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extreme f(x)=-3x^2+180x+190,0<= x<= 40

extreme points

f (x) = - 3 x^{2} + 180 x + 190, 0 \leq x \leq 40

Minimum (0, 190), Maximum (30, 2890), Minimum (40, 2590)

Solution steps

Find the critical points:

x = 0, x = 30, x = 40

Domain of

- 3 x^{2} + 180 x + 190, 0 \leq x \leq 40 : 0 \leq x \leq 40

Find intervals:

Increasing

: 0 < x < 30,

Decreasing

: 30 < x < 40

Plug the extreme point

x = 0

into

- 3 x^{2} + 180 x + 190 \Rightarrow y = 190

Minimum (0, 190)

Plug the extreme point

x = 30

into

- 3 x^{2} + 180 x + 190 \Rightarrow y = 2890

Maximum (30, 2890)

Plug the extreme point

x = 40

into

- 3 x^{2} + 180 x + 190 \Rightarrow y = 2590

Minimum (40, 2590)

Minimum (0, 190), Maximum (30, 2890), Minimum (40, 2590)

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

f (x) = - x^{3} - 6 x^{2} + 6

\frac{x ^{2} - 1}{x ^{2} + 4}

f (x, y) = y - x

11 x^{2} + 2 \cdot (- 25) x (1 - x) + 24 (1 - x)^{2}

What is the extreme f(x)=-3x^2+180x+190,0<= x<= 40 ?
The extreme f(x)=-3x^2+180x+190,0<= x<= 40 is Minimum(0,190),Maximum(30,2890),Minimum(40,2590)