Question

Consider the function h(x)=3-(5/x). The graph of h(x) is given. Does the MVT apply on the interval [-1,5]? Explain why or why not.

Answer 1

hint:

http://www.sosmath.com/calculus/diff/der11/der11.html

at the top it says

"The Mean Value Theorem is one of the most important theoretical tools in Calculus. It states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b)..."

Answer 2

ok so for interval [-1,5] the MVT does not apply since f(x) is not continuous at x=0, correct?

Answer 3

you nailed it

Answer 4

it's also not defined there

Answer 5

ok next question,
Does the MVT apply on the interval [1,5]?

Answer 6

yes because h(x) is continuous on [1,5] and differentiable on (1,5)

Answer 7

right.

Answer 8

correct

Answer 9

next question
Graphically, what does the MVT guarantee for the function on the interval [1,5]?

Answer 10

this one I am not sure

Answer 11

but how do I describe that in words?

Answer 12

is that the secant line will equal the tangent line?

Answer 13

are you given a set of choices? or are you just filling in a blank?

Answer 14

basically, MVT tells you that there is some point c whose tangent line will be parallel to the line ab over in the [a,b] interval

Answer 15

no my attachment is the worksheet I have to fill in. I have no key and I am doing my best to do it.

Answer 16

I believe the worksheet is design to introduce MVT and how it applies and how it doesn't apply from a graphical and algebraic approach

Answer 17

in math "words"
\[\text{for}~~[a,b]~~c\in[a,b] : a<c<b\therefore \vec{ab} ||\vec{c} \]

Answer 18

agreene is correct

here is the graph of h(x)=3-(5/x) with the points (1,-2) and (5,2) on it. These points represent the endpoints of the interval [1,5]

Answer 19

Draw a straight line through those two points. This line is the secant line (shown in blue).

The MVT says that there is at least one point in this interval where the slope of the tangent line is equal to the slope of the secant line (ie the tangent and secant lines are parallel). This tangent line is shown in green.

Answer 20

great that is what I drew on my paper

Answer 21

ok last question wants me to find the value(s) of c guaranteed for h(x) on the interval [1,5]
I found that
\[c=\sqrt{\frac{ 15 }{ 2 }}\]

Answer 22

thanks, I think I finished this....but most important I understood this a bit better

Answer 23

h(x) = 3-(5/x)

h ' (x) = ??

Answer 24

\[h'(x)=\frac{ 5 }{ x^2 }\]

Answer 25

the slope of the secant line is ______

Answer 26

it's +1, not -1

Answer 27

remember the MVT doesn't apply for [-1,5] because of x = 0

Answer 28

2/3 is incorrect

Answer 29

sorry, I will correct my paper

Answer 30

yep so 5/x^2 = 1 ---> x = ??

Answer 31

h'(c)=1

Answer 32

let me redo my c value

Answer 33

\[c=\sqrt{5}\]

Answer 34

you got it

Answer 35

thanks, I graph both tangent lines and they are very close graphically........

Answer 36

Thanks to both of you.......

Answer 37

you're welcome