Rolle and MVThm

Rolle's Theorem and the Mean Value Theorem

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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Rolle's Theorem

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 f is differentiable on the open interval (a, b),

and if f (a) = f (b), then

f '(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 (x) = 2x + 3, then f (c) = 2c + 3

Similarly, if f ' (x) = 2x + 3, then f '(c) = 2c + 3

So, when expressing f ' (c), just replace the variable (usually x -- with c).

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Example 1

Show that f (x) = 4x² – 20x + 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)= 8x – 20, so f '(c) = 8c – 20

f '(c) = 8c – 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/2a = 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 f is differentiable on the open interval (a, b), then

f '(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 differentiable on the open interval ( a, b ) before we can apply the theorem.

Also, c cannot be an endpoint: c is not equal to either a or b.

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Example 2

If f (x) = x³ – 8x – 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( 3c² – 8 ) = 39 becomes 3c² – 8 = 13

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

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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 x = 1 lies on [ 0, 2 ], the first condition -- that f (x) be continuous on [ a, b ] is not met. Therefore, there is no value of c on the interval ( 0, 2 ) that satisfies the conclusion of the theorem.

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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 not differentiable on (0, 2). Therefore both theorems do not apply.

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Practice

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) = 3x² + 2x + 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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Solutions

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 ]

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) = 3c² – 6c + 2 = 0

c = 1.57735, and c = 0.42265

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

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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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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) = 3x² + 2x + 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) = 6c + 2

So, 2(6c + 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) = – 2e^{– 2 c}

So, 3( – 2e^{– 2 c} ) = e^{– 6} – 1