What is the extreme f(x)=4sqrt(x^2+1)-3x ?

The extreme f(x)=4sqrt(x^2+1)-3x is Minimum(3/(sqrt(7)),sqrt(7))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=4sqrt(x^2+1)-3x

extreme points

f (x) = 4 x^{2} + 1 - 3 x

Minimum (\frac{3}{7}, 7)

Solution steps

f^{'} (x) = \frac{4 x}{x ^{2} + 1} - 3

Find intervals:

Decreasing

: - \infty < x < \frac{3}{7},

Increasing

: \frac{3}{7} < x < \infty

Plug

x = \frac{3}{7}

into

4 x^{2} + 1 - 3 x : 7

Minimum (\frac{3}{7}, 7)

f (x) = 75 - x^{2}

f (x) = \frac{x ^{3}}{3} - 3 x^{2} + 8 x - 2, 0 \leq x \leq 3

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

f (x) = ((45 - x) x - (10 + 24 x))

y = 7 + 5 x - 5 x^{2}

What is the extreme f(x)=4sqrt(x^2+1)-3x ?
The extreme f(x)=4sqrt(x^2+1)-3x is Minimum(3/(sqrt(7)),sqrt(7))