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Popular >

extreme 1/(t^3+2)

Solution

extreme points

\frac{1}{t ^{3} + 2}

Solution

Saddle (0, \frac{1}{2})

Solution steps

Find the critical points:

t = 0

Domain of

Find intervals:

Decreasing

Decreasing

Decreasing

: 0 < t < \infty

Plug the extreme point

x = 0

into

\frac{1}{t ^{3} + 2} \Rightarrow y = \frac{1}{2}

Saddle (0, \frac{1}{2})

Popular Examples

extreme f(x)= 6/7 (x^2-9)^{2/3}extreme points

f (x) = \frac{6}{7} (x^{2} - 9)^{\frac{2}{3}}

extreme f(x,y)=7xy+14x-x^2+2y^2 extreme points

f (x, y) = 7 x y + 14 x - x^{2} + 2 y^{2}

extreme 4x^2-8x^4 extreme points

4 x^{2} - 8 x^{4}

extreme f(x)=(x^2(x-1))/(x+2),x\ne 2 extreme points

f (x) = \frac{x ^{2} ( x - 1 )}{x + 2}, x \neq = 2

extreme f(x)=9x^3+21x^2+8x-3 extreme points

f (x) = 9 x^{3} + 21 x^{2} + 8 x - 3

Frequently Asked Questions (FAQ)

What is the extreme 1/(t^3+2) ?
The extreme 1/(t^3+2) is Saddle(0, 1/2)