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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

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

Solution steps

Find the critical points:

x = - 1, x = 0

Domain of

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

Find intervals:

Increasing

: - 1 < x < 0

Plug the extreme point

x = - 1

into

- x^{2} + 1 \Rightarrow y = 0

Minimum (- 1, 0)

Plug the extreme point

x = 0

into

- x^{2} + 1 \Rightarrow y = 1

Maximum (0, 1)

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

f (x) = x^{5} - x^{3} + 8, - 1 \leq x \leq 1

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

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

f (x) = 12 x + 6

f (x) = x (23 - 47 + 2 x) (\frac{47}{2} - x)

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