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### Course: AP®︎/College Calculus AB > Unit 5

Lesson 1: Using the mean value theorem- Mean value theorem
- Mean value theorem example: polynomial
- Mean value theorem example: square root function
- Using the mean value theorem
- Justification with the mean value theorem: table
- Justification with the mean value theorem: equation
- Establishing differentiability for MVT
- Justification with the mean value theorem
- Mean value theorem application
- Mean value theorem review

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# Mean value theorem example: square root function

Sal finds the number that satisfies the Mean value theorem for f(x)=√(4x-3) over the interval [1,3].

## Want to join the conversation?

- How did you end up getting a one in place of the f'(c)(7 votes)
- The "1" is a coincidence from the interval used in the video: 1 <=
**x**<= 3

Which makes it harder to follow the math, so I want to show how the "1" plugs back into the Mean Value Theorem (stick with me, this gets kind of long):

Here's how it works, the**x**values of 1 & 3

are based on the interval: 1 <=**x**<= 3.

We'll split f(x) = sqrt(4x-3) into smaller parts in this table to make the computation easier.`| x |4x - 3 | sqrt( 4x - 3 ) | f(x) | ( x, y ) |`

+-----+--------+----------------+-----------+----------+

| 1 |4*1 - 3 | sqrt( 4 - 3 ) | sqrt( 1 ) | ( 1, 1 ) |

| 3 |4*3 - 3 | sqrt( 12 - 3 ) | sqrt( 9 ) | ( 3, 3 ) |

Kind of a pain that**x**and**y**have the same value for this example, makes it harder to trace the math. But that is just the way this example problem in the video is set up. If we had a different function or a different interval it would be easier to see the difference.

So anyway, now we have two points for ( x, y ): (1,1) and (3,3).

Let's find the slope for interval 1 <=**x**<= 3.

Thinking about line equations, like y = mx + b

We just want the slope (the**m**part), which is works out like this:`(x1,y1) = ( 1, 1 )`

(x2,y2) = ( 3, 3 )

note: x1 from smallest x on interval 1 <= x <= 3

y1 from f(x1)

x2 from largest x on interval 1 <= x <= 3

y2 from f(x2)

y2 - y1 3 - 1 2

m = --------- = -------- = --- = 1

x2 - x1 3 - 1 2

So that is where we get the "1" from.

Now it is time to stuff the "1" into the Mean Value Theorem, which basically says:`There is at least a single value of c (may be more) such that:`

f(x2) - f(x1)

f'(c) = -------------- = 1

x2 - x1

Caution: The above is ONLY equal to "1" for this video's example problem,

(e.g. this specific function f(x) = sqrt(4x-3) and the interval 1 <=**x**<= 3).

For different intervals, like 10 <=**x**<= 100, we will*not*get "1" :-)

For different functions we will*probably*not get "1" (but we might, depends on the fnct and the interval).*Anyway*, now we can to solve f'(c) = 1.

Which looks like this: we will start with f(x) to find f'(x):`f(x) = sqrt( 4x - 3)`

note: since sqrt(x) = x^0.5 we can change as follows:

0.5

f(x) = ( 4x - 3)

note: using 0.5 instead of 1/2 for exponent because it looks cleaner in this "font".

(0.5 -1) d

f'(x) = 0.5 * ( 4x - 3 ) * --- ( 4x -3 )

dx

note: Power rule applied to ( 4x - 3) and

then chain rule to the inside part "4x -3"

-0.5

f'(x) = 0.5 * ( 4x - 3 ) * 4

note: the " * 4" on the right is the from chain

rule above, because:

d/dx 4x - 3 = d/dx 4x - d/dx 3

= 4 - 0

= 4

ok, moving on....

-0.5

f'(x) = 4 * 0.5 * ( 4x - 3 )

note: simplifying the 0.5(left) and 4 (from far right)

-0.5

f'(x) = 2 * ( 4x - 3 )

2

f'(x) = ------------------

sqrt( 4x - 3 )

note: simplify the 'negative exponent', e.g. a^-b = 1 / (a^b)

so ok to move (4x-3)^-0.5 down to the denominator.

Now we can solve for f'(c) = 1`2`

f'(c) = ------------------ = 1

sqrt( 4c - 3 )

note: next we'll multiply each side by the denominator...

2

sqrt( 4c - 3 ) ------------------ = 1 * sqrt( 4c - 3 )

sqrt( 4c - 3 )

2 = sqrt( 4c - 3 )

4 = 4c - 3

note: squaring each side to get rid of the sqrt()

4+3 = 4c - 3 + 3

note: add +3 to each side.

7 = 4c

7 * 1/4 = 4c * 1/4

multiply each side by 1/4

7/4 = c

Which is (more or less) how Sal got c = 7/4 in the video.

Which is a*really**long*way of saying: f'(7/4) = 1**Important**: the only reason we care about that is because of our "1",

the original coincidence from way up top.

- - - - - - - - - - - - - - - -**Different Example**x = [ 7, 21 ]

- - - - - - - - - - - - - - - -

So just for fun, let's try a different interval:`7 <= x <= 21`

| x | 4x - 3 | sqrt( 4x - 3 ) | f(x) | ( x, y )

+----+----------+-----------------+-------------+-----------

| 7 | 4*7 - 3 | sqrt( 28 - 3 ) | sqrt( 25 ) | ( 7, 5 )

| 21 | 4*21 - 3 | sqrt( 84 - 3 ) | sqrt( 81 ) | ( 21, 9 )

Finally :-) we have different x and y values.

Finding our slope y=mx + b style:

y2-y1 9 - 5 4

----- = ------ = ---- = 2/7

x2-x1 21 - 7 14

So, now we want f'(c) = 2/7

From above we already know f'(x):

2 2

f'(c) = -------------- = ----

sqrt( 4c - 3 ) 7

2

2 = ---- * sqrt( 4c - 3 )

7

14 = 2 * sqrt( 4c - 3 )

7 = sqrt( 4c - 3 )

49 = 4c - 3

49 + 3 = 4c

52 = 4c

52/4 = c

26/2 = c

13 = c

So, on interval 7 <=**x**<= 21 we get:

(edit: originally I wrote <= 12, which Michelle pointed out is wrong; fixed as <= 21)`2 f(12) - f(7) f(x2) - f(x1)`

f'( 13 ) = --- = ------------ = --------------- = f'(c)

7 12 - 7 x2 - x1

So, yeah, I think it is easier to follow with something other than "1".(28 votes)

- when finding f(3) and f(1) cant it also equal -3 and -1 because of the square

root??(3 votes)- It could, but the we'd have two y values for every x value. If that happens, we don't have a function anymore!

So mathematicians decided that, when taking square roots in a function, always use just the positive root (unless otherwise specified by a`-`

or`±`

in front of the root sign).

I hope this helps!(12 votes)

- What if you have a function over x? Does this mean you can't apply the Theorem because if you divide by x, you'd get undefined right?(2 votes)
- A function over x will have a removable discontinuity (like f(x) = [x(x+1)]/x) or a asymptote (like g(x) = (x+1)/x) in its graph in the point x = 0, thus it's not continuos at that point, and the Mean Value Theorem requires a closed interval where the function is continuos. You can still apply the Theorem in a function over x, but the interval can't be one that includes x = 0.(3 votes)

- This is my first year teaching calculus and I haven't done calculus in 20 years. I have come across a problem that asks is the M.V.T. applicable to f(x) = sqrt(x-3) on the interval [3,7]. The key I have says that because of the end point F(x) is not continuous on the interval so MVT can't be applied. But I have looked online and seen where some have said that a function can be considered continuous at an end point and differentiable at an end point. SO which is correct. I just want to get a better understanding so I can explain it to my students.(2 votes)
- Yeah, I agree that the MVT is applicable here.

The function is clearly differentiable over (3, 7] and thereby also continuous there, and since 𝑓(3) = lim_𝑥→3⁺ 𝑓(𝑥) it is also continuous in 𝑥 = 3.(2 votes)

- Hi! So I am taking Physics as well...how will the Mean Value Theorem help me in this class?(1 vote)
- It likely won't. The mean value theorem is an existence proof, and it's not constructive (it doesn't give you a formula to follow to find the point where f'(c)=[f(b)-f(a)]/[b-a] ), so it probably won't be helpful to find concrete solutions as you would need in physics.(4 votes)

- What if you get a value for c that is one of the numbers in the interval that is given? Would this be a possible c value?(1 vote)
- I'm not too sure what you mean. The point of the question is to find some x = c that is in the given interval (a, b) that satisfies the MVT. If you were referring to if c could be one of the endpoints (i.e., a or b), I am somewhat sure that c cannot equal a or b, as given in the definition of the MVT. Hope that I helped.(3 votes)

- in mean value theorem example ;square root function in video6:23..when we are finding f'x ,then how we can put 4 .

while if we are multiply with 4 then we should be also divide by 4(1 vote)- The 4 comes from the derivative of 4𝑥 − 3.

Remember the chain rule:

𝑓(𝑥) = 𝑔(ℎ(𝑥)) ⇒ 𝑓 '(𝑥) = 𝑔'(ℎ(𝑥)) ∙ ℎ'(𝑥)

In this case, ℎ(𝑥) = 4𝑥 − 3 ⇒ ℎ'(𝑥) = 4

and 𝑔(𝑥) = √𝑥 ⇒ 𝑔'(𝑥) = 1∕(2√𝑥)

⇒ 𝑔'(ℎ(𝑥)) = 1∕(2√(4𝑥 − 3))

Hence, 𝑓 '(𝑥) = 1∕(2√(4𝑥 − 3)) ∙ 4(2 votes)

- 5:45Shouldn't he check if 7/4 is an extraneous solution?(1 vote)
- It is possible that he didn't say that he checked it because it is assumed.(1 vote)

- How does knowing the mean value theorem help us with any future math problems?(1 vote)

## Video transcript

- [Voiceover] Let f of x
be equal to the square root of four x minus three,
and let c be the number that satisfies the mean
value theorem for f on the closed interval
between one and three, or one is less than or equal to x is less than or equal to three. What is c? So let's just remind ourselves
what it means for c to be the number that satisfies
the mean value theorem for f. This means that over this
interval, c is a point, x equals c is a point where
the slope of the tangent line at x equals c, so I
could write f prime of c, so that is the slope of the tangent line when x is equal to c, this is equal to the slope of the secant line that connects these two points. So this is going to be equal to, see the slope of the
secant line that connects the points three, f of
three, and one, f of one. So it's going to be f
of three minus f of one over three minus one. And if you wanted to think
about what this means visually, it would look something like this. So if this is our x axis and this is one, two, actually let me spread
it out a little bit more. One, two, and three. And so you have one, comma,
f of one, right over there, so that is at the point
one, comma, f of one, and we could evaluate that, actually, that's one, comma, one. So that's gonna be the
point one, comma, one. And then you have the point three, comma, let's see, you're gonna
have four times three is 12 minus three is nine, so it's
gonna be three, comma, three. So maybe it's right over
there, three, comma, three, and the curve might look
something like this, so it might look something like that, so if you think about
the slope of the line that connects these two points, so this line that
connects those two points, all the mean value theorem, I'm doing a different color. All the mean value
theorem tells us is that there's a point between
one and three where the slope of the tangent line
has the exact same slope. So if I were to eyeball it, it looks like it's right around there, although we are actually
going to solve for it. So, some point where the
slope of the tangent line is equal to the slope of the line that connects these two end points and their corresponding function values. So that is c, that would
be c right over there. So really we just have to solve this. So let's first just find
out what f prime of x is, and then we could substitute a c in there and then we can evaluate
this on the right hand side. So I'm gonna rewrite f of x, f of x is equal to, and I'm gonna write it as four x to the minus
three to the one half power. Makes it a little bit more
obvious that we can apply the power rule and the chain rule here. So f prime of x is going
to be the derivative of four x minus three to the one half with respect to four x minus three, so that is going to be
one half times four x minus three to the negative one half, and then we're gonna multiply that times the derivative of four x minus three with respect to x. Well, derivative of four x
with respect to x is just four, and the derivative of negative
three with respect to x, well that's just gonna be zero, so the derivative of four x minus three with respect to x is four. So times four. so f prime of x is equal to four times one half, which is two, over the square
root of four x minus three. Four x minus three to the one half would just be the square
root of four x minus three, but it's the negative one half, so we're gonna put it in the
denominator right over here. And so f prime of c, we could rewrite this as two over the square
root of four c minus three, and what is that going to be equal to? That is going to be equal to, let's see, f of three we already
figured out is three, f of one we already figured out is one, and so we get three minus
one over three minus one, well that's gonna be two over
two, which is equal to one. So there's some point
between one and three where the derivative at that point, the slope of the tangent
line, is equal to one. So let's see if we can solve
this thing right over here. We can multiply both
sides of this by four c by the square root of four c minus three, and so then we are going to get two is equal to the square
root of four c minus three. All I did is multiply both sides of this by square root of four c minus three to get rid of this in the denominator, and so let's see, now we
get rid of the radical, we can square both sides, and so actually let me just show that, so now we can square both sides, so we get four is equal
to four c minus three, add three to both sides,
seven is equal to four c, and then divide both sides by four. I'll go right here to do it, you're going to get c is
equal to seven fourths. C is equal to seven over four, which is equal to one and three fourths, or we could view this as 1.75. So actually, the c value
is a little bit closer, I hand drew this, it's closer
to about right over there on our diagram, actually
that looks pretty close, that actually looks pretty good, I just hand drew this curve
so it's definitely not exact. But anyway, hopefully that gives you a sense of what's going on here. We're just saying, hey,
the mean value theorem gives us some c where the
slope of the tangent line is the same as the slope of the line that connects one, f of
one, and three, f of three.