Mean value theorem problem. I know what the MVT is but I don't know how i'd use it to solve this problem.
for part a) I would take the derivative
okay i'l take it.
g ' (t) = 2 sin t * cos t - 3 g ' (t) = sin(2t) - 3
i got 2sin(t)cos(t)-3
the range of sin(x) is [-1, 1] so the range of sin(x) -3 is [ -1 -3 , 1 -3] = [ -4 , -2 ]
you can simplify 2 sin t cos t , using a trig identity
oh okay. that makes sense.
so you dont need MVT for part a)
g ' (t) < 0 for all t , therefore g(t) decreases in every interval in the domain
okay i'm following
do you see why g ' (t) < 0 for all t ?
yeah because it's range is -4 to -2 and both are less than 0
ok for part b) i think we are going to use MVT :)
okay so that is the f'(c) = f(b) - f(a) over b -a
\[\sin^2(t) - 3t = 5\] is given to us.
so do we plug the 5 into the MVT as the c?
i think we need to modify the equation first let f(x) = sin^2(t) -3t - 5 we want to solve f(x) = 0
sorry f(t) = sin^2(t) -3t -5
solutions to f(t) = 0 are solutions to the equation sin^2(t) -3t = 5
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.
lets use first intermediate value theorem
Find f(-Pi/2) = f(0) =
okay
working on it
for the first one i got -4 - 3pi/2 the second i got -5
f( -pi/2) = sin(-pi/2)^2 - 3(-pi/2) - 5 = 1 + 3*pi/2 - 5 = 3 pi/2 - 4 = .71238
but the initial equation was sin^2(t) -3t -5
shouldn't the 3pi/2 be subtracted?
there are two negatives there
oh sorry about that.
no problem :)
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: \[ \sin(2c)-3 = \frac{f(b)-f(a)}{b-a} \implies (b-a)(\sin(2c)-3) = f(b)-f(a) \]
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.
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
I think once you do part a, then it is easy to show that there can only be one root
Assuming there are two or more solutions leads to a contradiction because: \[ x_1<x_2\quad\text{ yet }\quad f(x_1)=f(x_2) \]
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
using f(t) = sin^2(t) -3t - 5
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.
Thank you two so very much.
and there can't be more than one solution since its a decreasing function
okay, that makes sense.
Also you can mention that f(t) is continuous and differentiable on any interval [a,b] , which satisfies the hypotheses of MVT
We showed it was decreasing. Which we can express mathmatically as: \[ a<b \implies f(b)<f(a) \]
This consequentially means: \[ a\neq b\implies f(b)\neq f(a) \]
right.
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