Rolle's Theorem and the Mean Value Theorem |

**Two Theorems on a Closed Interval**

This lesson covers 2 related theorems regarding the derivative of a continuous function over a closed interval. Each of them can be useful when we're looking for specific points on a closed interval -- such as a local maximum or minimum -- or a point where the tangent is parallel to a specific path.

Rolle's theorem can be very useful when dealing with polynomials for it enables us to find the critical numbers of a function. It also sets up the discovery of the Mean Value Theorem for Derivatives.

The usual descriptions of these two theorems look like a whole bunch of gobbledygook which no one can follow after a few words or terms -- so we'll decribe the situations in a picture and then state the gobbledygook.

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In street talk, Rolle's theorem says "if we have a chunk of a continuous function that begins and ends at the **same y-value** -- somewhere on that chunk, there's a **horizontal tangent**". Since the slope of the tangent is *f '*(*x*), and a horizontal line has slope = 0, there's a point on the chunk where the derivative is equal to 0.

Rolle's theorem makes it easy to locate the critical numbers of a function once we've found a pair of points with the same y-value. So, if we know that *f *(*x*) = 0 at *x* = 5 and at *x *= 9, then we know there's a critical number between 5 and 9. This is a property of continuous polynomial curves -- since there are no asymptotes or discontinuities. And, we're talking about a "*chunk*" of a continuous function with -- pay attention -- **equal ***y***-values at the endpoints**. Since the *y*-values are the same, and the curve has no holes, it either went straight across -- the contant function, or it went through a trough or crest of a wave -- otherwise known as a minimum or maximum.

**Note:** *c* cannot be an endpoint of the interval -- so *c cannot = a *or* b*.

Here's the formal statement of Rolle's Theorem.

If f(x) is a continuous function on closed interval [a, b]and and if
(c) = 0 for at least one value of c on (a, b). |

Since the *y*-values are the same at the endpoints of the interval (*a, b*), the slope of the "chord" joining the endpoints is = 0. So the theorem really says there's a point of tangency parallel to the "chord " joining the endpoints of the curve.

**Note:** We know that if ** f **(

Similarly, if *f** ' *(** x**) = 2

So, when expressing *f** '* (** c**), just replace the variable (usually

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**Example 1**

Show that *f *(*x*)* *= 4*x*² – 20*x* + 29 satisfies the hypothesis of Rolle's Theorem on [1, 4], then find all values of *c* such that *f '*(*c***)** = 0.

**Solution**

First, we show that *f *(*x*) satisfies the continuity and differentiable stipulations, then we show that the endpoint *y*-values are equal so that we can set *f '*(*c***)** = 0.

Since *f *(*x*) is a quadratic function, it is continuous and differentiable over all Real numbers.

* f* (1) = 13 =* **f *(4) so the endpoint *y*-values are equal.

* **f *' (*x*)= 8*x* – 20, so *f '*(*c***)*** *= 8*c* – 20

*f '*(*c***)** =* *8*c* – 20 = 0, so *c* = 5/2 = 2.5

since 2.5* *is on the interval (1, 4) and not an endpoint, *c* = 2.5.

We know that this parabola's vertex is at *h = – b*/2*a* = 20/8 = 5/2

so we have correctly found the *x*-value of the minimum on the interval.

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Now for the **Mean Value Theorem for Derivatives**, we simply rotate the images created by Rolle's Theorem, and come up with the fact that the "base" of the chunk doesn't have to be horizontal for there to be a point of parallel tangency on the interval.

In street talk, the **Mean Value Theorem for Derivatives** says "if we have a chunk of a continuous function -- then somewhere on that chunk, there's a **tangent parallel to the chord** formed by joining the endpoints of the chunk". Let's look at the pictures about Rolle's Theorem. Point ( *c*,* f *(*c*) ) is the point where the tangent is parallel to the "chord". If we just rotate each of the images, in any direction, that parallel tangent remains parallel -- only now, the slope of both the tangent and the chord is not 0 -- however, the lines remain paralle to each other.

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**Mean Value Theorem for Derivatives**

Notice that *f '(c)* which is the slope of the tangent

equals -- the slope of the line joining the endpoints.

If f(x) is a continuous function on closed interval [a, b]and
(c)( b – a ) = f (b ) – f (a ) for some c on (a, b). |

As with Rolle's Theorem, *f *(*x*) must be **continuous** **on** the closed interval **[ a, b ]** and

Also, *c* **cannot** be an endpoint: *c* is ** not equal** to either

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**Example 2**

If *f *(*x*)* = x³* – 8*x* – 5, show that *f* satisfies the hypothesis of the Mean Value Theorem on the interval [ 1, 4 ] and find *c* that satisfies the conclusion of the theorem.

**Solution**

Since *f* is a polynomial function of degree 3, it is continuous and differentiable over all Reals.

On [ 1, 4 ], | b – a = 3, |
f (4) = 27, |
f (1) = – 12, |
and f '(c) = 3c² – 8. |

So, 3( 3*c*² – 8 )* *= 39 becomes 3*c*² – 8 = 13

so

But since *c* must be on ( 1, 4 ) we reject the negative root.

so

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**Example 3**

Show that does not satisfy the conclusion of the Mean Value Theorem on the closed interval [ 0, 2 ].

**Solution**

Since ** f (x) is discontinuous at x = 1** (a vertical asymptote), and

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**Note** -- this example shows us it's important to **confirm the condition statements** before we try to find *c *in both theorems. If the conditions are not met, there is no *c* to look for!

The other condition stated in both theorems is that *f *(*x*) be **differentiable** on ( *a, b *). Visually, this means that the curve is such that there's a tangent to it at every point on the interval. If we go back to the image of the surfboard riding the wave -- the slope of that board being = the derivative of the curve's function, -- we're saying the board doesn't have to make any fancy-shmancy flips to ride the wave. It can do so smoothly.

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**Cusps** on a curve are **points** where the tangent slope goes through a **drastic change** -- usually from one infinity to the other. Cusps are "*pointy points*".

See how when our board rides **the outside of the wave** from left to right across points **P** or **Q**, -- we're going to need an airborne about-face flip to continue to ride the outside of the wave. The tangent slope goes from to over a single point. Since there is no tangent to the curve at the cusp point -- *f *(*x*) is **not differentiable ** at that point.

In surf talk -- the 2 conditions for both theorems are:

1) We can draw the wave over the interval (*a, b*) without picking up our pen -- and

2) We can ride the wave over the interval (*a, b*) without any fancy airborne flips.

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**Example 4**

Show that *f *(*x*) = |* x* – 1 | does not satisfy the conclusion of either Rolle's Theorem or the Mean Value Theorem on the interval [ 0, 2 ].

**Solution**

Though *f *(*x*) in this case is **continuous** on ( 0, 2 ), it is an **absolute value function** and goes through a **cusp at x = 1**, which is on the interval, so it is

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1) Verify that the function satisfies the 3 hypotheses (conditions) of **Rolle's Theorem** on the given interval. Then find *all* numbers *c* that satisfy the conclusion of Rolle's Theorem.

a) f (x) = x³ – 3x² + 2x + 5, [0, 2] |
b) | c) |

2) Verify that the function satisfies the 2 hypotheses (conditions) of the **Mean Value Theorem** on the given interval. Then find *all* numbers *c* that satisfy the conclusion of the Theorem.

a) *f *(*x*) = 3*x*² + 2*x* + 5,* ***[ –1, 1]** ....... b) *f *(*x*) =* e*^{ – 2 x} , **[ 0, 3 ]** ..... c)

3) A civil engineer is designing a "sound-blocking wall" for a housing development situated near a highway curve. The curve follows the rule

The diagram is a scaled down version of the situation. The engineer needs to find the coordinates of point **T** in order to know at what slope to build the wall for maximum sound-blocking effect. The wall must be parallel to the line **PQ**. (diagram)

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1) Verify that the function satisfies the 3 hypotheses (conditions) of **Rolle's Theorem** on the given interval. Then find *all* numbers *c* that satisfy the conclusion of Rolle's Theorem.

a) *f *(*x*) = *x*³ – 3*x*² + 2*x* + 5, **[ 0, 2 ]**

**Sol'n:** Since *f* is a polynomial function, it is continuous and differentiable on the Reals.

*f*(0) = 5 =* f*(2) so the endpoint *y*-values are equal.

*f *'(*c*) = 3*c*² – 6*c* + 2 = 0

* *

*c* = 1.57735, and *c** =* 0.42265

Since both belong to ( 0, 2 ), both solutions work.

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b)

**Sol'n:** Since *f* is a sine curve, it is continuous and differentiable on the Reals.

*f*( – 1) = 0 =* f*(1), so the endpoint *y*-values are equal.

therefore

Since, on the unit circle, *cos A = x-value*, and *x* = 0 at or

then

Therefore

Since both belong to ( – 1, 1), both solutions work.

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c)

**Sol'n:** Since *f* has a square root, it is continuous on* x* greater or = to – 6*.*

This function is differentiable on ( – 6, 0).

*f *( – 6) = 0 =* f* (0), so the endpoint *y*-values are equal.

Multiply through by , and solve to get *c* = – 4.

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2) Verify that the function satisfies the 2 hypotheses (conditions) of the **Mean Value Theorem** on the given interval. Then find *all* numbers *c* that satisfy the conclusion of the Theorem.

a) *f *(*x*) = 3*x*² + 2*x* + 5,* ***[ –1, 1]**

**Sol'n:** Since *f* is a polynomial function, it is continuous and differentiable on the Reals.

*f *( – 1) = 6, and *f* (1) = 10, and *b – a* = 1 – ( – 1) = 2.

*f *'(*c*) = 6*c* + 2

So, 2(6*c* + 2) = 4, which makes *c* = 0.

Since 0 is not an endpoint and is on the interval, the solution is *c* = 0.

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b) *f *(*x*) =* e*^{ – 2 x} , **[ 0, 3 ]**

This is an **exponential** function, therefore continuous and differentiable on all Reals.

*f *(0) = 1, and *f *(3) = *e*^{ – 6} , and *b – a* = 3 – (0) = 3.

*f *'(*c*) = – 2*e*^{ – 2 c}* *

So, 3( – 2*e*^{ – 2 c} ) =* **e*^{ – 6 } – 1

Now, we divide both sides by – 6 and take *ln* of both sides of the equation.

Since *ln e = 1*, we can ignore it. Divide both sides by *-2* and evaluate.

This value is on the interval so the solution is as shown.

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c)

Since *f* is **discontinuous with a vertical asymptote at x = 2, **which is

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3) A civil engineer is designing a "sound-blocking wall" for a housing development situated near a highway curve. The curve follows the rule

Find the coordinates of point **T**. (see diagram)

**Solution**

Since **P** is the vertex of *f*, at (2, 3) and

Since we move 36 units horizontally, **Q** is at .

**T** is the mean value point -- the point where the tangent is parallel to the chord.

So, since *b – a* = 36 , and *f *(*b*) – *f *(*a*) = 3/4, then 36* f '*(*c*) must = 3/4.

Now,

When we solve for *c*, we get *c* = 11 and *f* (11) = 27/8.

the coordinates of **T** are .

*(all content of the MathRoom Lessons **© Tammy the Tutor; 2004 - ).*