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I was navigating through math exercises about functions and I got with this question

If $f:[-1,1] \rightarrow \mathbb{R}$ defined by $f(x) = 12x^2 + 5x\sqrt{1-x^2} - 10$.

Give the range of $f(x)$ in the reals.

Any help will be appreciated

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Hint : If they exist, can you find the minimum and maximum points? – Euler....IS_ALIVE Oct 25 '12 at 21:24
Note that $f$ is only defined on $[-1,1]$, hence candidates for extrema are $-1$, $+1$ and zeroes of $f'$ in that interval. – Hagen von Eitzen Oct 25 '12 at 21:24
Well,I know the domain is -1,1 and the answer, it is -21/2, 5/2, but I am interested in how to solve it algebraically. Any other hints? – chubakueno Oct 25 '12 at 21:27
Hagen von Eitzen has given you the process. What do you get for $f(-1), f(1), f'(x)?$ – Ross Millikan Oct 25 '12 at 21:44
I did't know how to use derivatives, but thank you for the help – chubakueno Oct 26 '12 at 1:51
up vote 6 down vote accepted

Finding the derivative and solving $f'(x)=0$ may be unpleasant. So it is natural to let $x=\cos t$ where $0\le t\le \pi$. Our function is then $12\cos^2 t+5\cos t\sin t-10$. If we are in a calculus mood, differentiate.

But if we are in a trigonometric mood, we can use the identities $\cos 2t=2\cos^2 t-1$ and $\sin 2t=2\sin t\cos t$ to rewrite our expression as $6\cos 2t+\frac{5}{2}\sin 2t -4$, or equivalently $$\frac{13}{2}\left(\frac{12}{13}\cos 2t+\frac{5}{13}\sin 2t\right)-4.$$ Let $\varphi$ be an angle whose sine is $\frac{12}{13}$ and whose cosine is $\frac{5}{13}$. Then our function is $$\frac{13}{2}\sin(2t+\varphi) -4.$$

For $t$ in our interval, $\sin(2t+\varphi)$ has maximum value $1$ and minimum value $-1$. From this we see that the maximum and minimum values of $f(x)$ are $\dfrac{13}{2}-4$ and $-\dfrac{13}{2}-4$.

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+1 tres elegant! – Michael Joyce Oct 25 '12 at 23:18
Thank you! That was very simple and useful. +1 – chubakueno Oct 26 '12 at 1:51

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