What is the extreme f(x)=2x^2-10x+5,0<= x<= 7 ?

The extreme f(x)=2x^2-10x+5,0<= x<= 7 is Maximum(0,5),Minimum(5/2 ,-15/2),Maximum(7,33)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=2x^2-10x+5,0<= x<= 7

extreme

Maximum (0, 5), Minimum (\frac{5}{2}, - \frac{15}{2}), Maximum (7, 33)

Solution steps

Find the critical points:

x = 0, x = \frac{5}{2}, x = 7

Domain of

2 x^{2} - 10 x + 5, 0 \leq x \leq 7 : 0 \leq x \leq 7

Find intervals:

Decreasing

: 0 < x < \frac{5}{2},

Increasing

: \frac{5}{2} < x < 7

Plug the extreme point

x = 0

into

2 x^{2} - 10 x + 5 \Rightarrow y = 5

Maximum (0, 5)

Plug the extreme point

x = \frac{5}{2}

into

2 x^{2} - 10 x + 5 \Rightarrow y = - \frac{15}{2}

Minimum (\frac{5}{2}, - \frac{15}{2})

Plug the extreme point

x = 7

into

2 x^{2} - 10 x + 5 \Rightarrow y = 33

Maximum (7, 33)

Maximum (0, 5), Minimum (\frac{5}{2}, - \frac{15}{2}), Maximum (7, 33)

What is the extreme f(x)=2x^2-10x+5,0<= x<= 7 ?
The extreme f(x)=2x^2-10x+5,0<= x<= 7 is Maximum(0,5),Minimum(5/2 ,-15/2),Maximum(7,33)