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(a) Find the points of local maxima and minima, if any, of the following function defined in $0\leq x\leq 6$. $$x^3-6x^2+9x+15$$
(b) Integrate $$\int_{-\pi}^{\pi} x \cos x dx$$
asked in Calculus by Veteran (59.7k points) | 680 views

1 Answer

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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 $0$.
answered by Boss (11.3k points)
edited by
+8
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.
0
Do we need to consider endpoints always:For both local and Absolute extremums?
0
@Arjun Sir

then $cos x$ will be even function and $x cos x$ is an odd function.rt?
0
Yes, cos(x)  is an even function because cos(-x) = cos(x).
0
@ABKUNDAN  @Bikram  @akash.dinkar12

Do we need to consider endpoints always:For both local and Absolute extremums?
+1
yes, end points are given in the question.
0
for local maxima and minima also?
0
@Anurag end points will be considered only for 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?

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