Question

Mean Value Thereom Question. Question attached

Answer 1

I did the first part. Just need help with the second part

Answer 2

Never mind can u guys help me with this question. i already put it too many tries and i cant imput any more anwers to taht question. Please look at this one

Answer 3

Show that the equation has exactly one real root.
5x + cos x = 0

Answer 4

the first part it right. for the second part you have to solve
oh did you change the question?

Answer 5

Well since i didnt understand the 1st question do u mind explaining it. Sorry

Answer 6

I got that c = DNE but I guess thats wrong

Answer 7

\[\frac{\ln(5)-\ln(1)}{5-1}=\frac{1}{4}\ln(5)\] take the derivative get \(\frac{1}{x}\) set it equal to \(\frac{1}{4}\ln(5)\) and solve for \(x\)

Answer 8

if the function satisfies the hypothesis of the mvt then "\(c\)" must exist, that is what the theorem says

Answer 9

i got x = 4/ln(5)

Answer 10

in your case you get \(c=\frac{4}{\ln(5)}\) but if it is too late for that, we can do the next question as well

Answer 11

Thanks soo much and do u mind explaining first why we had to set the 2 equations equal to each other. Then we can move on to the nxt question

Answer 12

the theorem says that under the suitable hypotheses, that there is a \(c\) in \([a,b]\) with
\[f'(c)=\frac{f(b)-f(a)}{b-a}\]
now that is all very abstract, but you have a specific \(f\) and a specific \(a\) and \(b\)

Answer 13

in your case \(f(x)=\ln(x),f'(x)=\frac{1}{x}, a=1,b=5\)
so the theorem says there is a number \(c\in (1,b)\) with
\[\frac{1}{c}=\frac{\ln(5)-\ln(1)}{5-1}\]

Answer 14

the right hand side is \(\frac{1}{4}\ln(5)\) so solving for \(c\) gives \(c=\frac{4}{\ln(5)}\)
the theorem of course does not say "solve for \(c\)" or even really tell you how to find \(c\) it just says it exists
the problem just added the "find \(c\) " part

Answer 15

Oh ok i get it now :) Makes more sense this way. Thanks. If u dont mind lets move onto teh 2nd question

Answer 16

It was Show that the equation has exactly one real root.
5x + cos x = 0

Answer 17

as for the second question, pick some value of \(x\) for which \(5x+\cos(x)\) is negative
anything less than \(-1\) will do
then find some value for which is it positive, again anything will do
since it is continuous, and at some point is negative and at another point it is positive it must be zero somewhere

now to show it has only one zero:
then take the derivative and note that it is always positive
this tells you the function is always increasing

those two facts taken together show that there is exactly one zero

Answer 18

When i take the derivative I get 5 - sinx. But dont we have to set taht equal to 0? I get that it DNE tahts y im confused

Answer 19

i guess what i wrote was not clear, or at least the second part.
your derivative is right, it is \(f'(x)=5-\sin(x)\)
the point of taking the derivative is to show that \(f(x)=5x+\cos(x)\) is a strictly increasing functions
if the derivative of a function is always positive, then the function is always increasing
is it clear that \(5-\sin(x)>0\) for all \(x\)?

Answer 20

Yes it it now

Answer 21

ok then i hope the entire "proof" makes sense,
it is negative, then positive, so it must be zero somewhere and it cannot cross the \(x\) axis twice because it is strictly increasing

Answer 22

Yes i get it now. Thank you for a thorough explanation. it makes perfect sense to me now :D

Answer 23

Haha got a 100 on my hw :) Thanks b/c those 2 q's I was stuck on.

Answer 24

yw