en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme t^4-4t^3+15

Solution

extreme points

t^{4} - 4 t^{3} + 15

Solution

Saddle (0, 15), Minimum (3, - 12)

Solution steps

f^{'} (t) = 4 t^{3} - 12 t^{2}

Find intervals:

Decreasing

: - \infty < t < 0,

Decreasing

: 0 < t < 3,

Increasing

: 3 < t < \infty

Plug

t = 0

into

t^{4} - 4 t^{3} + 15 : 15

Saddle (0, 15)

Plug

t = 3

into

t^{4} - 4 t^{3} + 15 : - 12

Minimum (3, - 12)

Saddle (0, 15), Minimum (3, - 12)

Popular Examples

extreme y=x(20-2x)(16-x)extreme points

y = x (20 - 2 x) (16 - x)

extreme y=e^x+1 extreme points

y = e^{x} + 1

extreme f(x)=x^3-6sqrt(x)+6 extreme points

f (x) = x^{3} - 6 x + 6

extreme f(x)=12x^3-24x^2+12x extreme points

f (x) = 12 x^{3} - 24 x^{2} + 12 x

extreme f(x)=8x^{5/3}+\sqrt[3]{x}extreme points

Frequently Asked Questions (FAQ)

What is the extreme t^4-4t^3+15 ?
The extreme t^4-4t^3+15 is Saddle(0,15),Minimum(3,-12)