What is the y=(x^4)/4+1/(8x^2),1<= x<= 2 ?

The y=(x^4)/4+1/(8x^2),1<= x<= 2 is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y=(x^4)/4+1/(8x^2),1<= x<= 2

y = \frac{x ^{4}}{4} + \frac{1}{8 x ^{2}}, 1 \leq x \leq 2

Domain : 1 \leq x \leq 2

Range : \frac{3}{8} \leq f (x) \leq \frac{129}{32}

Intercepts : None

Asymptotes : None

Extreme Points : Minimum (1, \frac{3}{8}), Maximum (2, \frac{129}{32})

Interval Notation

Domain : [1, 2]

Range : [\frac{3}{8}, \frac{129}{32}]

Solution steps

Domain of

\frac{x ^{4}}{4} + \frac{1}{8 x ^{2}}, 1 \leq x \leq 2 : 1 \leq x \leq 2

Range of

\frac{x ^{4}}{4} + \frac{1}{8 x ^{2}}, 1 \leq x \leq 2 : \frac{3}{8} \leq f (x) \leq \frac{129}{32}

Axis interception points of

\frac{x ^{4}}{4} + \frac{1}{8 x ^{2}}, 1 \leq x \leq 2 :

None

Asymptotes of

\frac{x ^{4}}{4} + \frac{1}{8 x ^{2}}, 1 \leq x \leq 2 :

None

Extreme Points of

\frac{x ^{4}}{4} + \frac{1}{8 x ^{2}}, 1 \leq x \leq 2 :

Minimum

(1, \frac{3}{8}),

Maximum

(2, \frac{129}{32})