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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 $$\int_{-\pi}^{\pi} x \cos x dx$$
in Calculus | 1.2k views

(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 $0$.

edited
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If anyone is not clear for (b)

cos(-x) = cos x, so -xcos(-x) = -xcosx, making xcosx an odd function. So, integral from $-\pi$ to 0 will be negative of the integral from 0 to $\pi$ making the sum 0.
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Do we need to consider endpoints always:For both local and Absolute extremums?
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@Arjun Sir

then $cos x$ will be even function and $x cos x$ is an odd function.rt?
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Yes, cos(x)  is an even function because cos(-x) = cos(x).
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@ABKUNDAN  @Bikram  @akash.dinkar12

Do we need to consider endpoints always:For both local and Absolute extremums?
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yes, end points are given in the question.
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for local maxima and minima also?
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@Anurag end points will be considered only for absolute extremums
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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.

Someone explain this //

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Why are we considering the end points also in local maxima/minima? They should be considered only when we are trying to find out the global maxima/minima, isnt it?
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@jatin khachane 1

that is because x lies between 0 and 6

So, 0 and 6 contains some extreme point i.e. critical point and we know critical point contains some local minima or local maxima atleast

@Somoshree Datta 5

no, that will be local, not global

+1

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.

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.

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