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Mathematics 51 Online
OpenStudy (anonymous):

Mean value theorem problem. I know what the MVT is but I don't know how i'd use it to solve this problem.

OpenStudy (anonymous):

OpenStudy (perl):

for part a) I would take the derivative

OpenStudy (anonymous):

okay i'l take it.

OpenStudy (perl):

g ' (t) = 2 sin t * cos t - 3 g ' (t) = sin(2t) - 3

OpenStudy (anonymous):

i got 2sin(t)cos(t)-3

OpenStudy (perl):

the range of sin(x) is [-1, 1] so the range of sin(x) -3 is [ -1 -3 , 1 -3] = [ -4 , -2 ]

OpenStudy (perl):

you can simplify 2 sin t cos t , using a trig identity

OpenStudy (anonymous):

oh okay. that makes sense.

OpenStudy (perl):

so you dont need MVT for part a)

OpenStudy (perl):

g ' (t) < 0 for all t , therefore g(t) decreases in every interval in the domain

OpenStudy (anonymous):

okay i'm following

OpenStudy (perl):

do you see why g ' (t) < 0 for all t ?

OpenStudy (anonymous):

yeah because it's range is -4 to -2 and both are less than 0

OpenStudy (perl):

ok for part b) i think we are going to use MVT :)

OpenStudy (anonymous):

okay so that is the f'(c) = f(b) - f(a) over b -a

OpenStudy (anonymous):

\[\sin^2(t) - 3t = 5\] is given to us.

OpenStudy (anonymous):

so do we plug the 5 into the MVT as the c?

OpenStudy (perl):

i think we need to modify the equation first let f(x) = sin^2(t) -3t - 5 we want to solve f(x) = 0

OpenStudy (perl):

sorry f(t) = sin^2(t) -3t -5

OpenStudy (perl):

solutions to f(t) = 0 are solutions to the equation sin^2(t) -3t = 5

OpenStudy (anonymous):

got it, so I'm not sure how to solve for t, considered quadtratic formula, but i don't think it's in the right form to do that.

OpenStudy (perl):

lets use first intermediate value theorem

OpenStudy (perl):

Find f(-Pi/2) = f(0) =

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

working on it

OpenStudy (anonymous):

for the first one i got -4 - 3pi/2 the second i got -5

OpenStudy (perl):

f( -pi/2) = sin(-pi/2)^2 - 3(-pi/2) - 5 = 1 + 3*pi/2 - 5 = 3 pi/2 - 4 = .71238

OpenStudy (anonymous):

but the initial equation was sin^2(t) -3t -5

OpenStudy (anonymous):

shouldn't the 3pi/2 be subtracted?

OpenStudy (perl):

there are two negatives there

OpenStudy (anonymous):

oh sorry about that.

OpenStudy (perl):

no problem :)

OpenStudy (anonymous):

For the first part... if you want to be more rigorous... We pick an arbitrary interval \([a,b]\). After we differentiate, we can say there is a \(c\), such that: \[ \sin(2c)-3 = \frac{f(b)-f(a)}{b-a} \implies (b-a)(\sin(2c)-3) = f(b)-f(a) \]

OpenStudy (anonymous):

Then: \[ \sin(2c)\leq 1\implies \sin(2c)-3\leq -2 \]And since \(a<b\implies 0<b-a\) A positive times a negative is negative..\[ f(b)-f(a) < 0 \implies f(b)<f(a) \]That is how I would prove it is always decreasing.

OpenStudy (perl):

and that uses MVT :) since (b-a) >0 and sin(2c) -3 < 0 then (b-a) (sin (2c) -3 ) < 0 and therefore f(b) - f(a) < 0 , which implies f(b) < f(a) , so f is decreasing

OpenStudy (anonymous):

I think once you do part a, then it is easy to show that there can only be one root

OpenStudy (anonymous):

Assuming there are two or more solutions leads to a contradiction because: \[ x_1<x_2\quad\text{ yet }\quad f(x_1)=f(x_2) \]

OpenStudy (perl):

since f(-pi/2) > 0 and f(0) < 0 By intermediate value theorem there exists c such that f(c) = 0 on the interval (-pi/2, 0). so we know there exists at least one root. if we assume there is more than one root we get a contradiction

OpenStudy (perl):

using f(t) = sin^2(t) -3t - 5

OpenStudy (anonymous):

Assuming there is no solutions is strange because:\[ \lim_{x\to-\infty}f(x) = -\infty\\ \lim_{x\to\infty}f(x) = \infty\\ \]So by the IVT, it has to hit \(5\) at some point because it is continuous.

OpenStudy (anonymous):

Thank you two so very much.

OpenStudy (perl):

and there can't be more than one solution since its a decreasing function

OpenStudy (anonymous):

okay, that makes sense.

OpenStudy (perl):

Also you can mention that f(t) is continuous and differentiable on any interval [a,b] , which satisfies the hypotheses of MVT

OpenStudy (anonymous):

We showed it was decreasing. Which we can express mathmatically as: \[ a<b \implies f(b)<f(a) \]

OpenStudy (anonymous):

This consequentially means: \[ a\neq b\implies f(b)\neq f(a) \]

OpenStudy (anonymous):

right.

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