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Find the local maximum and minimum values of the following: $t+ \cos t$

Note: This is part 6th of a 6 part question, split as 6 separate questions here.

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1 Answer

  • Second derivative test: Suppose $f$ is continuous on an open interval that contains
  • (i)If $f'(c)=0\; and\; f''(c) >0$ then $f$ has a local minimum at $c$.
  • (ii)If $f'(c)=0\;and \; f''(c)<0$ then $f$ has a local maximum at $c$
$f(t)=t+\cos t$
Step 1:
$f'(t)=1-\sin t$
At the turning points $f'(t)=0$
$f'(t)=0=>1-\sin t=0=>\sin t=1$
=>$ t=n \pi +(-1)^n \large\frac{\pi}{2}$
Step 2:
When $t=n \pi +(-1)^n \large\frac{\pi}{2}$
$f(t)=n \pi +(-1)^n \large\frac{\pi}{2}$
The stationary points are at $(n \pi+(-1)^n \large\frac{\pi}{2}$$, n \pi (-1)^n \large\frac{\pi}{2} )$$ n \in z$
Step 3:
$f''(t)=-\cos t$
When $t=n \pi +(-1)^n \large\frac{\pi}{2}$$,f^u (t)=0$ => the second derivative test fails
We consider the change in sign of the first derivative on either side of such values .
As $ 1 > | \sin t |, 1- \sin t$ is positive
$\therefore $ there are no local maxima /minima


answered Aug 1, 2013 by meena.p
edited Aug 1, 2013 by meena.p