What is the range of f(x)=2x^2-1/x ?

The range of f(x)=2x^2-1/x is -infinity <f(x)<infinity

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range of f(x)=2x^2-1/x

range

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

- \infty < f (x) < \infty

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

2 x^{2} - \frac{1}{x} : x < 0 or x > 0

Extreme Points of

2 x^{2} - \frac{1}{x} :

Minimum

Find the range for the interval

0 < x < \infty : - \infty < f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

- \infty < f (x) < \infty

\frac{8 - x ^{3}}{2 x ^{2}}

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

2 x - 5 ln (4 x + 2)

f (x) = \frac{2 x - 5}{x - 3}

f (x) = \frac{4 x}{4 x - 5}

What is the range of f(x)=2x^2-1/x ?
The range of f(x)=2x^2-1/x is -infinity <f(x)<infinity