What is the extreme f(x)=2x^2-4x+5,0<= x<= 7 ?

The extreme f(x)=2x^2-4x+5,0<= x<= 7 is Maximum(0,5),Minimum(1,3),Maximum(7,75)

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extreme f(x)=2x^2-4x+5,0<= x<= 7

extreme points

f (x) = 2 x^{2} - 4 x + 5, 0 \leq x \leq 7

Maximum (0, 5), Minimum (1, 3), Maximum (7, 75)

Solution steps

Find the critical points:

x = 0, x = 1, x = 7

Domain of

2 x^{2} - 4 x + 5, 0 \leq x \leq 7 : 0 \leq x \leq 7

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 7

Plug the extreme point

x = 0

into

2 x^{2} - 4 x + 5 \Rightarrow y = 5

Maximum (0, 5)

Plug the extreme point

x = 1

into

2 x^{2} - 4 x + 5 \Rightarrow y = 3

Minimum (1, 3)

Plug the extreme point

x = 7

into

2 x^{2} - 4 x + 5 \Rightarrow y = 75

Maximum (7, 75)

Maximum (0, 5), Minimum (1, 3), Maximum (7, 75)

(5.4) (4.3)

f (x) = \frac{10 - 3 x}{2}

f (x) = x + x + 10

f (x) = x^{2} - 6 x - 16

\frac{ln ( x - 1 )}{x - 1}

What is the extreme f(x)=2x^2-4x+5,0<= x<= 7 ?
The extreme f(x)=2x^2-4x+5,0<= x<= 7 is Maximum(0,5),Minimum(1,3),Maximum(7,75)