What is the extreme y=x^{4/7}(11/4+x) ?

The extreme y=x^{4/7}(11/4+x) is Maximum(-1, 7/4),Minimum(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme y=x^{4/7}(11/4+x)

extreme

Maximum (- 1, \frac{7}{4}), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{11 ( x + 1 )}{7 x ^{\frac{3}{7}}}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

x^{\frac{4}{7}} (\frac{11}{4} + x) : \frac{7}{4}

Maximum (- 1, \frac{7}{4})

Plug

x = 0

into

x^{\frac{4}{7}} (\frac{11}{4} + x) : 0

Minimum (0, 0)

Maximum (- 1, \frac{7}{4}), Minimum (0, 0)

What is the extreme y=x^{4/7}(11/4+x) ?
The extreme y=x^{4/7}(11/4+x) is Maximum(-1, 7/4),Minimum(0,0)