Question 1

How did you end up getting a one in place of the f'(c)

Accepted Answer

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".

Question 2

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

Accepted Answer

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!

Question 3

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.

Accepted Answer

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.

Question 4

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?

Accepted Answer

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.

Question 5

Hi! So I am taking Physics as well...how will the Mean Value Theorem help me in this class?

Accepted Answer

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.

AP®︎/College Calculus BC

Course: AP®︎/College Calculus BC > Unit 5

Mean value theorem example: square root function

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