Question
Find the intervals in which f(x) is increasing or decreasing:
$\text{f}(\text{x})=\sin\text{x}(1+\cos\text{x}),0<\text{x}<\frac{\pi}{2}$
$\text{f}(\text{x})=\sin\text{x}(1+\cos\text{x}),0<\text{x}<\frac{\pi}{2}$
✓
Answer
Consider the function,
$\text{f}(\text{x})=\sin\text{x}(1+\cos\text{x}),0<\text{x}<\frac{\pi}{2}$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}+\sin\text{x}(-\sin\text{x})+\cos\text{x}(\cos\text{x})$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}-\sin^2\text{x}+\cos\text{x}(\cos\text{x})$
$\Rightarrow\text{f}(\text{x})=\cos\text{x}+\big(\cos^2\text{x}-1\big)+\cos^2\text{x}$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}+2\cos^2\text{x}-1$
$\Rightarrow\text{f}'(\text{x})=2\cos^2\text{x}+\cos\text{x}-1$
$\Rightarrow\text{f}'(\text{x})=2\cos^2\text{x}+2\cos\text{x}-\cos\text{x}-1$
$\Rightarrow\text{f}'(\text{x})=2\cos\text{x}(\cos\text{x}+1)-1(\cos\text{x}+1)$
$\Rightarrow\text{f}'(\text{x})=(2\cos\text{x}-1)(\cos\text{x}+1)$
For f(x) to be increasing, we must have,
$\text{f}'(\text{x})>0$
$\Rightarrow\text{f}'(\text{x})=(2\cos\text{x}-1)(\cos\text{x}+1)>0$
$\Rightarrow0<\text{x}<\frac{\pi}{3}$
$\Rightarrow\text{x}\in\Big(0,\frac{\pi}{3}\Big)$
So, f(x) is increasing in $\Big(0,\frac{\pi}{3}\Big)$
For f(x) to be decreasing, we must have,
$\text{f}'(\text{x})<0$
$\Rightarrow\text{f}'(\text{x})=(2\cos\text{x}-1)(\cos\text{x}+1)<0$
$\Rightarrow\frac{\pi}{3}<\text{x}<\frac{\pi}{3}$
$\Rightarrow\text{x}\in\Big(\frac{\pi}{3},\frac{\pi}{2}\Big)$
So, f(x) is decreasing in $\Big(\frac{\pi}{3},\frac{\pi}{2}\Big)$
$\text{f}(\text{x})=\sin\text{x}(1+\cos\text{x}),0<\text{x}<\frac{\pi}{2}$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}+\sin\text{x}(-\sin\text{x})+\cos\text{x}(\cos\text{x})$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}-\sin^2\text{x}+\cos\text{x}(\cos\text{x})$
$\Rightarrow\text{f}(\text{x})=\cos\text{x}+\big(\cos^2\text{x}-1\big)+\cos^2\text{x}$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}+2\cos^2\text{x}-1$
$\Rightarrow\text{f}'(\text{x})=2\cos^2\text{x}+\cos\text{x}-1$
$\Rightarrow\text{f}'(\text{x})=2\cos^2\text{x}+2\cos\text{x}-\cos\text{x}-1$
$\Rightarrow\text{f}'(\text{x})=2\cos\text{x}(\cos\text{x}+1)-1(\cos\text{x}+1)$
$\Rightarrow\text{f}'(\text{x})=(2\cos\text{x}-1)(\cos\text{x}+1)$
For f(x) to be increasing, we must have,
$\text{f}'(\text{x})>0$
$\Rightarrow\text{f}'(\text{x})=(2\cos\text{x}-1)(\cos\text{x}+1)>0$
$\Rightarrow0<\text{x}<\frac{\pi}{3}$
$\Rightarrow\text{x}\in\Big(0,\frac{\pi}{3}\Big)$
So, f(x) is increasing in $\Big(0,\frac{\pi}{3}\Big)$
For f(x) to be decreasing, we must have,
$\text{f}'(\text{x})<0$
$\Rightarrow\text{f}'(\text{x})=(2\cos\text{x}-1)(\cos\text{x}+1)<0$
$\Rightarrow\frac{\pi}{3}<\text{x}<\frac{\pi}{3}$
$\Rightarrow\text{x}\in\Big(\frac{\pi}{3},\frac{\pi}{2}\Big)$
So, f(x) is decreasing in $\Big(\frac{\pi}{3},\frac{\pi}{2}\Big)$
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