Find the min/max values of $f(t)=\cos(t)+t \sin(t)$ for $t \in [0, \pi/2]$

Question

I have to find the minimum and maximum values of $f(t)$ for the specified interval for $t$. I considered using derivatives or substracting a common factor $\sqrt{1+t^2}$ in $f(t)$, so that $f(t)=\sqrt{1+t^2}\sin(t+\alpha)$ and I get only one trigonometric function to evaluate. But then I'm not sure what to do next. Any help would be appreciated.

Answer 1

$f'(t)=-\sin t+t\cos t+\sin t=t\cos t\geq 0$ for all $t \in [0,\frac {\pi} 2]$. Hence the function is increasing. Its maximum value is attained at $\frac {\pi} 2$ and the minimum value at $0$. Hence the minimum value is $1$ and the maximum value is $\frac {\pi} 2$.

Answer 2

We have $f'(t)= - \sin t + \sin t + t \cos t= t \cos t$. Then setting this to zero we have $t=0$ or $\cos t=0$. It is easy to show then that the only solutions for $t$ in the interval $[0,\pi/2]$ are $t=0$ and $t=\pi/2$. You need then only check if these are max/mins.