# Another range problem

Find the range of the function defined by f(x)=($x^{4}$ - $\sqrt{2}$x + 2)/($x^{4}$ - $\sqrt{2}$x + 1).

Note by Nishant Sharma
6 years, 5 months ago

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I got the lower bound. But don't know how to get the upper bound. Additionally can anyone help me to find the roots of the denominator of the given function f(x)(if they exist) and if they don't then how ?

- 6 years, 5 months ago

Letting x^4 - (rt2)x + 1 = y, you get f(x) = 1 + 1/y. Some work gives you y >= 1/4, giving you a maximum of 5.

- 6 years, 5 months ago

That is what I could not get. Since the denominator function is increasing and decreasing in certain intervals, so I couldn't figure out how to obtain upper bound.

- 6 years, 5 months ago

Use f'(x) to find out intervals in which the function is increasing or decreasing. You don't need the denominator in that case.

- 6 years, 5 months ago

here nishant ; ..... i think i could help you with only the answer but not the procedure how to solve it...if you don't mind i am here posting the link from where i could find answer of this question....hope this answer(only) helps you ......

- 6 years, 5 months ago

That was fine. Understood. Did you see my recent post ? I mean roots of denominator ?

- 6 years, 5 months ago

but i could not find it....is it in the discussion forum...?...then give me the heading that you gave...

- 6 years, 5 months ago

Split the function into $1+\frac{1}{(x^4-\sqrt{2}x+1)}$

Since it is easy to see that the lower bound will be greater than 1 but never equal to one. Now for upper bound we can find the local maxima.

$f'(x)=-\frac{4x^3-\sqrt{2}}{(x^4-\sqrt{2}x+1)^2}$

Equating $f'(x)=0$

We get $x=\frac{1}{\sqrt{2}}$

We can check whether it is the local maxima or not using the second derivative test(checked it).

Now substituting for $x=\frac{1}{\sqrt{2}}$ in $f(x)$ We have $f(\frac{1}{\sqrt{2}})=5$.

So range is$(1,5]$.

- 6 years, 5 months ago

That was nicely explained. Got it.

- 6 years, 5 months ago