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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

Maximum (- 1, 0), Minimum (- \frac{1}{2}, ln (\frac{3}{4})), Maximum (1, ln (3))

Solution steps

Find the critical points:

x = - 1, x = - \frac{1}{2}, x = 1

Domain of

ln (x^{2} + x + 1), - 1 \leq x \leq 1 : - 1 \leq x \leq 1

Find intervals:

Decreasing

: - 1 < x < - \frac{1}{2},

Increasing

: - \frac{1}{2} < x < 1

Plug the extreme point

x = - 1

into

ln (x^{2} + x + 1) \Rightarrow y = 0

Maximum (- 1, 0)

Plug the extreme point

x = - \frac{1}{2}

into

ln (x^{2} + x + 1) \Rightarrow y = ln (\frac{3}{4})

Minimum (- \frac{1}{2}, ln (\frac{3}{4}))

Plug the extreme point

x = 1

into

ln (x^{2} + x + 1) \Rightarrow y = ln (3)

Maximum (1, ln (3))

Maximum (- 1, 0), Minimum (- \frac{1}{2}, ln (\frac{3}{4})), Maximum (1, ln (3))

f (x) = x^{2} + x y + y^{2} - 28 y + 261

\frac{- x + 3}{x ^{2} + 14 x + 49}

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

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

f (x) = - 2 cos (2 x)

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