What is the extreme f(x)=x^2-2x-3,0<= x<= 1 ?

The extreme f(x)=x^2-2x-3,0<= x<= 1 is Maximum(0,-3),Minimum(1,-4)

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extreme f(x)=x^2-2x-3,0<= x<= 1

extreme points

f (x) = x^{2} - 2 x - 3, 0 \leq x \leq 1

Maximum (0, - 3), Minimum (1, - 4)

Solution steps

Find the critical points:

x = 0, x = 1

Domain of

x^{2} - 2 x - 3, 0 \leq x \leq 1 : 0 \leq x \leq 1

Find intervals:

Decreasing

: 0 < x < 1

Plug the extreme point

x = 0

into

x^{2} - 2 x - 3 \Rightarrow y = - 3

Maximum (0, - 3)

Plug the extreme point

x = 1

into

x^{2} - 2 x - 3 \Rightarrow y = - 4

Minimum (1, - 4)

Maximum (0, - 3), Minimum (1, - 4)

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

f (x) = 7 x + ln (x)

y = x^{\frac{2}{7}} (x^{2} - 5)

y = - t u (t - 1) + t u (t - 2)

f (x) = 2 x^{2} + 2 y^{2} - 8 x + 16 y

What is the extreme f(x)=x^2-2x-3,0<= x<= 1 ?
The extreme f(x)=x^2-2x-3,0<= x<= 1 is Maximum(0,-3),Minimum(1,-4)