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Mathematics 15 Online
OpenStudy (hdrager):

Verify that the function g(x)= 3x/x+7 satisfies the hypothesis of the Mean Value Theorem on the interval [-1,2]. Then find all the numbers c that satisfy the conclusion of MVT.

OpenStudy (royalranger):

Do you know what the Mean value theorem states?

OpenStudy (hdrager):

yeah but idk how to apply

OpenStudy (royalranger):

is this function continuous at every point of the closed interval [-1,2]?

OpenStudy (hdrager):

everywhere except x=-7

OpenStudy (hdrager):

there's a hole there

OpenStudy (hdrager):

oh so it is continuous on the interval

OpenStudy (royalranger):

So it is continuous on the interval [-1,2]. Which means it is differentiable on (-1,2). So find the derivavtive of the function.

OpenStudy (royalranger):

derivative*

OpenStudy (hdrager):

would the derivative just be 3?

OpenStudy (royalranger):

Is this your function? \[\frac{ 3x }{ x+7 }\]

OpenStudy (hdrager):

yes

OpenStudy (royalranger):

Have you learned the quotient rule yet?

OpenStudy (hdrager):

yeah

OpenStudy (hdrager):

ooooh oops

OpenStudy (royalranger):

I suggest you do it by hand to practice, but if you are already proficient in calculating the derivative, this could save time. Plus it shows the steps. http://www.derivative-calculator.net/

OpenStudy (hdrager):

\[\frac{ 21 }{ x^2+14x+49 }\]

OpenStudy (hdrager):

i tried it by hand

OpenStudy (royalranger):

Plug your function into the website i linked above, it will show you all the steps.

OpenStudy (hdrager):

i got it right

OpenStudy (hdrager):

they just didn't factor the denomiator

OpenStudy (hdrager):

so what do I do next

OpenStudy (royalranger):

So now that you have that. Find: \[\frac{ f(b)- f(a) }{ b-a }\]

OpenStudy (hdrager):

what was the point in taking the derivative of the original function then?

OpenStudy (royalranger):

Because after you find the value of \[\frac{ f(b)-f(a) }{ b-a }\] you will set that equal to the derivative and solve for x.

OpenStudy (hdrager):

also i got 15/6

OpenStudy (hdrager):

so then I set 15/6 = to the derivative?

OpenStudy (royalranger):

yep

OpenStudy (hdrager):

ahh okay I'm confused on how to solve here

OpenStudy (hdrager):

for x

OpenStudy (royalranger):

you could multiply both sides by 1/21 to get rid of the 21 in the numerator.

OpenStudy (hdrager):

okay so then I have (x+7)^2=15/126

OpenStudy (hdrager):

I unfactored it bc it seemed easier

OpenStudy (hdrager):

oops i messed up

OpenStudy (hdrager):

it would = 315/6

OpenStudy (hdrager):

then what should I do?

OpenStudy (royalranger):

move the fraction other side so your function is equal to zero.

OpenStudy (hdrager):

okay then what

OpenStudy (royalranger):

use the quadratic equation to find x

OpenStudy (hdrager):

wait so i would be using x^2+14x-3.5 for the quadratic formula?

OpenStudy (hdrager):

i just want to make sure before I go at it

OpenStudy (hdrager):

or should I stick with fractions?

OpenStudy (royalranger):

Alright so this whole time i have just been trusting that you have done all of the steps correctly without actually doing all of the work.

OpenStudy (royalranger):

i just did: \[\frac{ f(b)-f(a) }{ b-a }\] and i got 1/9

OpenStudy (hdrager):

wait how

OpenStudy (royalranger):

do you know what f(b) - f(a) means?

OpenStudy (hdrager):

yeah you plug in the points from the given interval

OpenStudy (royalranger):

\[\frac{ 3(2) }{ (2) +7 } -\frac{ 3(-1) }{ (-1)+7 } = \frac{ 6 }{ 9 } - \frac{ -3 }{ 6 } = \frac{ 1 }{ 3 }\] and 2 -(-1) = 3 therefore (1/3)/3 = (1/9)

OpenStudy (hdrager):

isn't it 2+1 in the denominator

OpenStudy (royalranger):

yes, which is 3 just like i posted above.

OpenStudy (hdrager):

\[x=-7\pm \sqrt{26}\]

OpenStudy (hdrager):

??

OpenStudy (hdrager):

@royalranger

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