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asked in Mathematical Logic by Boss (5.8k points)   | 109 views
Iam getting Max = $\frac{9}{8}$ and min = $0$
maximum match but minimum doesn't match the answer which has been given

max=1.125    //9/8=1.125

min = -2
how did u get max?
Yes, it has to be $-2$, i calculated min at $x=2\pi$, which is wrong.
Wait a minute. Writing answer!

1 Answer

+4 votes
Best answer

$\color{olive}{f(x) = \sin(x) + \cos(2x)}$

$\color{olive}{f'(x) = \cos(x) - 2\sin(2x)}$

$\color{olive}{f'(x) = \cos(x) - 4\sin(x)\cos(x)}$

$\color{olive}{f'(x) = \cos(x)(1-  4\sin(x))}$

For minima and maxima, $f'(x) = 0$

$f'(x) = \cos(x)(1-  4\sin(x)) = 0$

$\cos(x) = 0$ (or) $1 - 4\sin(x) = 0$

$x = \frac{\pi}{2},\frac{3\pi}{2}$ (or) $\sin(x) = \frac{1}{4} \Rightarrow x = \sin^{-1}(\frac{1}{4})$

Now, the proper method for finding minima and maxima is to check whether $f''(x) \gt 0$ or $f''(x) \lt 0$, but here as we are required to find both minima and maxima, i will directly put values of x and find out.

When $x = \frac{\pi}{2}$, $f(x) = 1 + (-1) = 0$

When $x = \frac{3\pi}{2}$, $f(x) = \sin(\frac{3\pi}{2}) + \cos(3\pi) = -1 + (-1) = -2$

$\sin(x) = \frac{1}{4}$, $\cos(x) = \frac{\sqrt{15}}{4}$, $\sin(2x) = 2\sin(x)\cos(x) = 2*\frac{1}{4}*\frac{\sqrt{15}}{4} = \frac{\sqrt{15}}{8}$ and $\cos(2x) = \frac{\sqrt{64-15}}{8} = \frac{7}{8}$

When, $x = \sin^{-1}(\frac{1}{4})$, $f(x) = \sin(x) + \cos(2x) = \frac{1}{4} + \frac{7}{8} = \frac{2}{8} + \frac{7}{8} = \frac{9}{8}$

Hence, $\color{navy}{f(x)_{max} = \frac{9}{8}}$ and $\color{navy}{f(x)_{min} = -2}$

answered by Veteran (24.7k points)  
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