What is the range of x^2ln(x) ?

The range of x^2ln(x) is f(x)>=-1/(2e)

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range of x^2ln(x)

range

x^{2} ln (x)

f (x) \geq - \frac{1}{2 e}

Interval Notation

[- \frac{1}{2 e}, \infty)

Solution steps

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

Domain of

x^{2} ln (x) : x > 0

Extreme Points of

x^{2} ln (x) :

Minimum

(\frac{1}{e}, - \frac{1}{2 e})

Find the range for the interval

0 < x < \infty : - \frac{1}{2 e} \leq f (x) < \infty

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

f (x) \geq - \frac{1}{2 e}

\frac{- x ^{2} - 6 x - 8}{x ^{2} - 2 x - 8}

f (x) = \frac{4}{- x - 2} + 2

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

f (x) = 4 x^{2} - 10 x + 3

y = \frac{1}{4} x - 6