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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+8 votes

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

+17 votes

Best answer

(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 $0$.

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 $0$.

+18

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.

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.

+1

@ABKUNDAN @Bikram @akash.dinkar12

Do we need to consider endpoints always:For both local and Absolute extremums?

Do we need to consider endpoints always:For both local and Absolute extremums?

0

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 //

0

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?

0

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

chk this https://gateoverflow.in/41/gate2012-9

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:

- $f'(x)=0;$
- $f'(x)$ 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*.

Source: https://mathinsight.org/local_minima_maxima_refresher

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

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