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# 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
3 years, 8 months ago

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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$$(1,5]$$. · 3 years, 8 months ago

That was nicely explained. Got it. · 3 years, 8 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 ...... · 3 years, 8 months ago

That was fine. Understood. Did you see my recent post ? I mean roots of denominator ? · 3 years, 8 months ago

but i could not find it....is it in the discussion forum...?...then give me the heading that you gave... · 3 years, 8 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. · 3 years, 8 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. · 3 years, 8 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. · 3 years, 8 months ago