What is the extreme f(x)=(x^3+3)/(2x^2-6x-4) ?

The extreme f(x)=(x^3+3)/(2x^2-6x-4) is Maximum(0.70192…,-0.46301…),Minimum(6.95933…,6.65360…)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(x^3+3)/(2x^2-6x-4)

extreme points

f (x) = \frac{x ^{3} + 3}{2 x ^{2} - 6 x - 4}

Maximum (0.70192 \dots, - 0.46301 \dots), Minimum (6.95933 \dots, 6.65360 \dots)

Solution steps

Find the critical points:

x \approx 0.70192 \dots, x = 6.95933 \dots

Domain of

\frac{x ^{3} + 3}{2 x ^{2} - 6 x - 4} : x < \frac{3 - 17}{2} or \frac{3 - 17}{2} < x < \frac{3 + 17}{2} or x > \frac{3 + 17}{2}

Find intervals:

Increasing

: - \infty < x < \frac{3 - 17}{2},

Increasing

: \frac{3 - 17}{2} < x < 0.70192 \dots,

Decreasing

: 0.70192 \dots < x < \frac{3 + 17}{2},

Decreasing

: \frac{3 + 17}{2} < x < 6.95933 \dots,

Increasing

: 6.95933 \dots < x < \infty

Plug the extreme point

x = 0.70192 \dots

into

\frac{x ^{3} + 3}{2 x ^{2} - 6 x - 4} \Rightarrow y = - 0.46301 \dots

Maximum (0.70192 \dots, - 0.46301 \dots)

Plug the extreme point

x = 6.95933 \dots

into

\frac{x ^{3} + 3}{2 x ^{2} - 6 x - 4} \Rightarrow y = 6.65360 \dots

Minimum (6.95933 \dots, 6.65360 \dots)

Maximum (0.70192 \dots, - 0.46301 \dots), Minimum (6.95933 \dots, 6.65360 \dots)

f (x) = \frac{x ^{3} - 2}{( x - 1 ) ^{2}}

f (x) = \frac{3}{x + 4}

f (x, y) = ((x + 9)^{2} - 1) (x^{2} - 16) + 200 y^{2} + 2

f (x, y) = 5 x y - 7 x^{2} - y^{2} + 3 x - 6 y + 2

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

What is the extreme f(x)=(x^3+3)/(2x^2-6x-4) ?
The extreme f(x)=(x^3+3)/(2x^2-6x-4) is Maximum(0.70192…,-0.46301…),Minimum(6.95933…,6.65360…)