How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for $F(x)= (x^2)(lnx)$ ?

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How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for $F(x)= (x^2)(lnx)$ ?

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Answer 1 · 2017-01-03T10:23:21

By analysing $F'(x)$ over the domain of $F(x): x>0$
$F(x)$ has an absolute minimum at $x=1/sqrt(e)$
$F(x)$ is decreasing for $0< x<1/sqrt(e)$ and increasing for $x>1/sqrt(e)$

Explanation:

$F(x) = x^2lnx$

$F'(x) = x^2.!/x + lnx*2x$ (Product Rule)

$= x+2x*lnx$

For local extrema $F'(x) =0$

Thus: $x+2xlnx = 0$

$x(1+2lnx)=0$

$x=0$ or $lnx=-1/2$

Since $F(x)$ is defined for $x>0$ our local extrema will be at:

$lnx=-1/2 -> x=e^(-1/2) = 1/sqrt(e) ~= 0.60653$

Hence:
$F(x)$ has an absolute minimum at $x=1/sqrt(e)$
$F(x)$ is decreasing for $x<1/sqrt(e)$ and increasing for $x>1/sqrt(e)$

This is shown by the graph of $F(x)$ below:

graph{x^2*lnx [-0.542, 2.877, -0.449, 1.26]}

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How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for $F(x)= (x^2)(lnx)$ ?

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Explanation:

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How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for F(x)= (x^2)(lnx)F(x)= (x^2)(lnx)?

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Explanation:

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How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for $F(x)= (x^2)(lnx)$ ?