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Calculus1 18 Online
OpenStudy (anonymous):

b) Find the value(s) of c that satisfy the conclusion of the MVT for the function f(x) =sqrt(x-1) on the interval [1; 3]. c) How many real roots does the equation x^5 + 2x + 3 = 0 have?

OpenStudy (anonymous):

first, do you know what the mean value theorem says?

OpenStudy (anonymous):

basically an equation will have a tangent line equal to the average slope of the line

OpenStudy (anonymous):

so first lets find the slope of the line in equation

OpenStudy (anonymous):

so (0-sqrt2)/1-3

OpenStudy (anonymous):

yup so that is \[\sqrt{2}/2\]

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

now lets take the derivative of your function

OpenStudy (anonymous):

(.5)(x-1)^(-.5)

OpenStudy (anonymous):

so 1/sqrtx-1

OpenStudy (anonymous):

1/2(ssqrtx-1)

OpenStudy (anonymous):

great, so now solve for x when the derivative equals sqrt(2)/2

OpenStudy (anonymous):

so plug sqrt2/2 into f'(x)?

OpenStudy (anonymous):

f'(sqrt/2)??

OpenStudy (anonymous):

not exactly, you are looking for what x value will guve you a derivative of sqrt(2)/2

OpenStudy (anonymous):

so solve\[\sqrt{2}/2=1/(2\sqrt{x-1})\]

OpenStudy (anonymous):

ok, then what? while i have your attention, haha

OpenStudy (anonymous):

then you have your answer

OpenStudy (anonymous):

oh i see, so my x value once solves for is my value "c"

OpenStudy (anonymous):

then what about the real roots equation, i feel like its just 5, but i don't really know how to prove it

OpenStudy (anonymous):

first, tell me what math course you are in because there are a few different ways to go about this problem depending on what course this is in

OpenStudy (anonymous):

calc 1

OpenStudy (anonymous):

ok. lets think about this, what happens to f(x) as x approaches infinity?

OpenStudy (anonymous):

it goes toward infinity

OpenStudy (anonymous):

and what about as f(x) approaches negative infinity

OpenStudy (anonymous):

negative infinity

OpenStudy (anonymous):

ok, good, do you know what an inflection point is?

OpenStudy (anonymous):

f''

OpenStudy (anonymous):

f''(x)=0

OpenStudy (anonymous):

yeah so that is when the derivative is going to change sign (become positive or negative)

OpenStudy (anonymous):

can you tell me what the first derivative of this function is?

OpenStudy (anonymous):

5x^4 + 2

OpenStudy (anonymous):

i have to leave for class in 5 minutes

OpenStudy (anonymous):

does that equal zero for any real x value? (we are almost done)

OpenStudy (anonymous):

yeah i think it does

OpenStudy (anonymous):

since the derivative is always positive here, we know that there arent any dips in the function so it can only cross that x axis once. This means that there is only one real root

OpenStudy (anonymous):

x^4 is always postive so 5x^4 +2 is always positive

OpenStudy (anonymous):

ohhhhhh

OpenStudy (anonymous):

wow

OpenStudy (anonymous):

if you look at this http://www.wolframalpha.com/input/?i=x%5E5%2B2x%2B3 you will see that it only crosses the x axis once

OpenStudy (anonymous):

so we can use the derivative to see that the function is always increasing

OpenStudy (anonymous):

right right, i understand so basically it was asking how many times does it cross the x axis, that makes so much more sense

OpenStudy (anonymous):

i get where u are going with the +/- infinity too now, ahha

OpenStudy (anonymous):

if it is always increasing from negative infinity to infinity, it will only cross the x axis once

OpenStudy (anonymous):

Thank you very much

OpenStudy (anonymous):

no prob, enjoy class

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