verify that the hypothesis of the Mean value theorem are satisfied on the given interval and findall values of c that satisfy theconclusion of the theorem. f(x)=x^2+x [-4,6]

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verify that the hypothesis of the Mean value theorem are satisfied on the given interval and findall values of c that satisfy theconclusion of the theorem. f(x)=x^2+x [-4,6]

Mathematics
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Do you understand how the Meal-Value Theorem works? :O
hm.. i'll try to do this one myself and ask for help if i still can't get the answer :)

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|dw:1358281444723:dw|Ok but here's a brief explanation of the concept c: just in case it helps.
|dw:1358281503651:dw|
Remember the formula for finding the slope of a line between 2 points? (a secant line) If we use the notation from the graph,\[\large m=\frac{f(b)-f(a)}{b-a}\] In order for the function to be continuous, this slope must be equal to the slope of a TANGENT line somewhere in the middle.\[\large \frac{f(b)-f(a)}{b-a}=f'(c)\] A derivative at a particular point represents the slope of a tangent line.
ok, i need help... i dont know how to do this.. haha i thought i did but i dont. so, please continueee :)
So in this problem we're starting with a secant line, passing through x=-4 and x=6. Let's start by finding the slope \(m\) of that line.\[\large m=\frac{f(6)-f(-4)}{6-(-4)}\]
\[\large m=\frac{\color{orangered}{f(6)}-\color{cornflowerblue}{f(-4)}}{6-(-4)}\] \[\large m=\frac{(\color{orangered}{6^2+6})-(\color{cornflowerblue}{(-4)^2-4})}{6+4}\]
Understand how those got plugged in?
YES :)
Simplifying that down, I think we end up withhhhhhhhh 3ish.
yes. i got 3.
So the theorem says - In order for this function to be continuous from -4 to 6, there must be a TANGENT line somewhere between those points that has this slope of 3.
Let's first find f'(x).
What'd u get? :O
2x+1 :)
k cool :D
So then,\[\large f'(c)=2c+1\]And we're claiming, that by the Mean-Value Theorem this tangent line at x=c has a slope of 3. It has to happen somewhere between -4 and 6 in order for the function to be continuous. So let's set our f'(c) equal to 3 and solve for c. \[\large 2c+1=3\]
c=1
When you get an answer, verify whether or not it is BETWEEN -4 and 6. Because that will help to verify that we didn't make a mistake somewhere.
since it's 1, it is between -4 and 6 :)
Yay team c:
yes. but is that the final answer? i thought we were supposed to use the mean value theorem.. which says: if F is continuous on [z,b], and if F is an anti-derivative of F on [a,b], then \[\int\limits_{a}^{b}f(x)dx=f(b)-f(a)\].
oops, i meant [a,b]..
Oh crap maybe i confused it with the intermediate value theorem... sec.. thinking :3
ahh darn. haha ok ;p
hmm no it doesn't appear i did... hmmmm
The thing you posted is the FTC (Fundamental Theorem of Calculus, Part 2).\[\large \int\limits_a^b f(x)dx=F(b)-F(a)\] I'm not quite sure what that has to do with the mean value theorem D: hmmm
oh sorryyy yea thta IS the fundamental theorem of calculus.. ha but the mean value theorem of integrals says : if f is conb], then there exists at least one number x* in [a,b] such that \[\int\limits_{a}^{b}f(x)dx=f(x*)(b-a).\]
if f is continuous on [a,b] i meant.
Hmm crap I dunno D: I'm familiar with seeing it that way.
aw darn it.. is there anyone you could ask help for? because that formula is what my teacher uses... :'(
@hartnn @hba @campbell_st Maybe one of these fellas know what you're talking about c:
ahh thankss :)
oh nvermind! we got the correct answer. so i';lljust do it the way @zepdrix did it! :D thank you for comming anyway! :)
ohh..okay :)

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