Question

Find the minimum and maximum of the function

y=\dfrac{x^4+x^2+5}{(x^2+1)^2}

Collected in the board: Function

Steven Zheng posted 6 hours ago

Answer

y=\dfrac{x^4+x^2+5}{(x^2+1)^2}

=\dfrac{x^4+2x^2+1-x^2-1+5}{(x^2+1)^2}

=\dfrac{(x^2+1)^2-(x^2+1)+5}{(x^2+1)^2}

=1-\dfrac{1}{x^2+1}+\dfrac{5}{(x^2+1)^2}

Let

t = \dfrac{1}{x^2+1}

Then

0< t\leq 1

and

y = 5t^2-t+1 = 5(t-\dfrac{1}{10} )^2+\dfrac{19}{20}

When t = \dfrac{1}{10} , y_{min} = \dfrac{19}{20}

Solving the equation t =\dfrac{1}{x^2+1} = \dfrac{1}{10} gives x = \pm3

Since 0< t\leq 1 , y has maximun values when t = 1

y_{max} = 5 -1+1 = 5

Solving the equation t =\dfrac{1}{x^2+1} = 1 gives x = 0

In summary, the function y has minimum value \dfrac{19}{20} when x=\pm3 and maximun value 5 when x=0

Steven Zheng posted 6 hours ago

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