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  1. Find the points of local maxima and minima, if any, of the following function defined in $0\leq x\leq 6$. $$x^3-6x^2+9x+15$$
  2. Integrate $$\int_{-\pi}^{\pi} x \cos x dx$$
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(a)$f(x)=x^{3}-6x^{2}+9x+15$

so $f'(x)=3x^{2}-12x+9=0 \Rightarrow x=1,3$

Now $f''(x)=6x-12$

$f''(1) < 0,$ so $x = 1$ is point of local maxima, $f''(3) > 0,$ so $x = 3$ is point of local minima.

Also the end points $0$ and $6$ are critical points. $0$ is point of local minima, because it is to the left of $x = 1$ (which is point of maxima). Similarly $x = 6$ is point of local maxima.

(b) Since $x \cos x$ is an odd function, by the properties of definite integration, answer is $0$.
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Also the end points 0 and 6 are critical points.

End points are not critical points. A point x is a critical point if it satisfies either of the following conditions:

  1. $f'(x)=0;$
  2. $f'(x)$ is undefined.

Source: https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-analytical-applications-new/ab-5-2/v/minima-maxima-and-critical-points

Here we do check for endpoints because they are included in the domain (closed interval). 

The endpoints require separate treatment: There is the auxiliary point to just $t_{0}$ the right of the left endpoint a, and the auxiliary point $t_{n}$ just to the left of the right endpoint b:

  • At the left endpoint a, if  $f'(t_{0})< 0$ (so $f'$ is decreasing to the right of a) then a is a local maximum.
  • At the left endpoint a, if $f'(t_{0})> 0$ (so $f'$ is increasing to the right of a) then a is a local minimum.
  • At the right endpoint b, if $f'(t_{n})< 0$ (so $f'$ is decreasing as b is approached from the left) then b is a local minimum.
  • At the right endpoint b, if $f'(t_{n})> 0$ (so $f'$ is increasing as b is approached from the left) then b is a local maximum.

Source: https://mathinsight.org/local_minima_maxima_refresher

Now what if end-points are not included in the domain (open interval)?

See this: https://gateoverflow.in/20002/tifr2011-a-4

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Thanks @Nirmal .This also cleared my doubt of 2012 question of maxima and minima.
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$f””(x)=6x-12$

So If we check the value at 1,3 first

at x=0 double derivative of f(x) = -12 Hence Shouldn’t this be the point of Local Minima?

at x=6 double derivative of f(x) = 24 >0 Hence Shouldn’t this be the point of Local Minima?
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