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Rolle’s Theorem – Proof & Problems | Schooling

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Rolle’s Theorem – Proof & Problems

Rolle’s Theorem – Proof & Problems

What is Rolle’s Theorem?

Rolle’s Theorem is just one of the necessary Calculations in theorems. It is reported the pursuing:

Let f(x) fulfill the following conditions:

Functionality f is carry on on the shut interval (a,b)

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

f(a) = f(b)

Then there must be a number c these kinds of that a < c < b and if(c) = 0. Now, this Theorem is only exact if the three conditions are fulfilled. Though the third condition is easy to check. And a lot of people struggle to see if the formula 1^st1

st

and formula 2^nd2

nd

conditions are fulfilled. Let us take a look at some methods that can use to see if a function is continuous and differentiable.

How to Tell if a Function is Continuous

The best way to see if a function is continuous then is by finding the values of xx that make the role not clear. Then we see if the xx value is within the closed interval of [a,b]. Whether it is, then the function is discontinuous if it is not. Then it is continuous. Let us look at an example.

On Question No.1: Show that function f is continuous on the closed interval [1,7].

Rolle’s Theorem

Formulae No. 1

Equation No. 1: Continuous Polynomial Question part .1

If you look at the function carefully, one will notice that this is a polynomial. Generally, the standard sets of rules of a polynomial are:

Formulae No. 2

Continuous Polynomial Question part .2

Equation No.1: Continuous Polynomial Question part .2

It means that there are no xx values. In which the function is undefined. In other words, the procedure is continuous from – infty−∞ to infty∞. If that is the case, it must also be continuous on these closed intervals [1,7]. Okay, so that was easy. Let us look at a more challenging question.

Question 2: Is this the function continuous on the closed interval of [0,2]?

Formulae No. 3

Equation 2: Discontinuous Rational Question part.1

Equation 2: Discontinuous Rational Question part.1

It is a fraction. And a fraction already has a problem if the denominator is equal to 0. It is because we cannot divide by 0. So there is a relation to be an xx value in each function which is undefined. Let us set the denominator equal to 0 to see what that is:

Formulae No. 4

Equation 2: Discontinuous Rational Question part.2

Equation 2: Discontinuous Rational Question part.2

Adding both sides by one gives:

Formulae No. 5

Equation 2: Discontinuous Rational Question part.3

Equation 2: Discontinuous Rational Question part.3

Cube rooting both sides gives us to have

Formulae No. 6

Equation 2: Discontinuous Rational Question part.4

Equation 2: Discontinuous Rational Question part.4

The function is undefined at x = 1. Now is this one is within the interval of [0,2]? We can see that it is. So the function is not continuous at the interval of [0,2]. Now that we are familiar with this continuity let us take a look at what it means to be differentiable.

What Does it Mean to be Differentiable?

A function is differentiable at the intervals of (a, b). Suppose its derivative exists within that interval that we have to do here. Take the result and check the domain. Suppose an xx value within the interval of (a, b) makes the derivative undefined. We must say that the function f(x) is not differentiable. Let us look at an example.

Question No. 3: Is the function f(x) = frac1x^2 + 1f(x)=

x

2

+1

1

​ differentiable on the open interval of (-1,1)?

First, we need to take the derivative. Doing so gives us:

Equation 3: Differentiable Rational part.1

Equation 3: Differentiable Rational part.1

Simplifying to gives us:

Formulae No. 7

Equation 3: Differentiable Rational part.2

Equation 3: Differentiable Rational part.2

Now we are going to see when the derivative is undefined. Again, the best way to do sum is by setting the denominator equal to 0. Doing so gives us:

Formulae No. 8

Equation 3: Differentiable Rational part.3

Equation 3: Differentiable Rational function.3

Square rooting on both sides of the equation gives to

Equation 3: Differentiable Rational part.4

Equation 3: Differentiable Rational function.4

On Subtracting both sides by one offer:

Formulae No. 9

Equation 3: Differentiable Rational pt.5

Equation 3: Differentiable Rational part.5

Square rooting both sides again gives to an x value:

Equation 3: Differentiable Rational part.6

Equation 3: Differentiable Rational function.6

Notice that this is an imaginary number. So this is not a valid x value. That means there are no x values. That makes the derivative undefined. So we can end that the function is differentiable at the intervals of (-1,1). Let us do a different example.

Question No. 4: Is the function f(x) = frac2xx^2 – 9f(x)=

x

2

−9

2x

​ differentiable on the open interval of (-4, 4)?

Again, taking the derivative that gives us:

Equation 4: Non-Differentiable Rational part.1

Equation 4: Non-Differentiable Rational part.1

On Simplifying the numerator that gives:

Equation 4: Non-Differentiable Rational part.2

Equation 4: Non-Differentiable Rational part.2

Again, let us set the denominator equal to 0. To see when the function is undefined.

Equation 4: Non-Differentiable Rational part.3

Equation 4: Non-Differentiable Rational part.3

On Square rooting both sides gives:

Equation 4: Non-Differentiable Rational part.4

Equation 4: Non-Differentiable Rational part.4

Adding the numerical 9 to both sides gives to

Equation 4: Non-Differentiable Rational part.5

Equation 4: Non-Differentiable Rational part.5

On Square rooting, both sides again give two x-values:

Equation 4: Non-Differentiable Rational part.6

Equation 4: Non-Differentiable Rational part.6

So, these values make the derivative not clear. Notice these x values are within the open interval of (-4, 4). Hence, we end that the function f(x) is not differentiable at the break (-4, 4).

Rolle’s Theorem Proof

Now we are familiar with the conditions of Rolle’s theorem, let us prove the theorem itself. This theorem says that if a function f(x) satisfies all three conditions. There must a number c such at a < c < b and if(c) = 0. We can also be showing that this is always true if we prove that it is true for each of these cases:

A function with only a constant at an interval of [a,b]

A position with a maximum at an interval of [a,b]

A process with a minimum at an interval of [a,b]

Let us look at each case separately.

A Function with only a Constant

Find the function f (x) = k within the interval of [a,b]. Where k is a constant. Notice that differentiating the function is gives:

Equation 5: Constant function in case 1

Equation 5: Constant process in case 1

It is interesting because the derivative is always 0. Since f'(x) = 0 for all x, then we can take any number within the interval of [a,b] as c. We finds a number c such that a < c < b and f'(c) = 0. That was very easy. Let us look at the second case.

A Function with a Maximum

Find some number k in the interval of [a,b] such that f (k)> f(a) = f(b). If f(x) is continuous, then there is a utmost at level c. Given that we all know f(x) is differentiable (from the 2nd problem). Then we know that f'(x) exists. Given that f'(x) exists and there is a highest inside the interval of [a,b]. Then we know that f'(c) = inside of the interval of [a,b]. There will exists a quantity c these types of that a< c < b and f'(c) = 0. Finally, let us look at the third case.

A Function with a Minimum

Find some number k in the interval of [a,b}. Like that f (k) < f(a) = f(b). Whether f(x) is continuos then means there will exists minimum at C point. Since we all know f(x) is differentiable from the 2nd condition. We know that f'(x) exists. Since f'(x) exists and there is a minimum within the interval of [a,b]. Then we know that f'(c) = 0 within the interval of [a,b]. In other words, it exists a number c such that a < c < b and f'(c) = 0.

Since in all cases are true, then Rolle’s Theorem is proved. Now let us do some examples of Rolle’s Theorem.

The Rolle’s Theorem Examples

Question No. 5: Showing that the equation has exactly one real root.

Equation 6: One Real root Polynomial part.1

Equation 6: One Real root Polynomial part.1

Now from this function, one should realize that this is a 7^th7.

The degree polynomial, therefore it has a total of 7 roots. Since the question proves it has only one real root then the six other roots must complex. How do we show this? Well, notice that.

Equation 6: One Real root Polynomial part.2

Equation 6: One Real root Polynomial part.2

Besides, we can see that

Equation 6: One Real root Polynomial part.3

Equation 6: One Real root Polynomial part.3

Ago -9 < 0 < 1 and f(x) is a polynomial is continuous everywhere, then the character Value Theorem proves that there must a number c such that -1 < c < 0, and f(c) = 0. In other words, a process must have at least one real root. However, we need to show f(x) has exactly one real root, not at least one. So what do we do?

We do something is called contradictory proof.

Let us guess f(x) has at least two real roots. It means we can find at least two x values of (call it a, b) where f(a) = 0 and f(b) = 0. Now here is the interesting part. If one did not notice, we satisfied all the conditions for Rolle’s Theorem. First, f(x) is a polynomial. So it is continuous and different anywhere. It satisfies the 1s and 2nd conditions. Lastly, we should know that f(a) = f(b) because they both equal to 0. So this satisfies the 3rd condition. So what does Rolle’s Theorem tell on this? It says that more be a number c such that a < c < b and f'(c) = 0.

But looking at this one if we take the derivative of f(x), then we get:
Equation 6: One Real root Polynomial part.4
Equation 6: One Real root Polynomial part.4

If one looks closely, the derivative can never equal 0. It is because the exponents are even. That always leads to favorable terms. One also has to add 3 to it, so it becomes even more advantageous. So, therefore, we are particular that it is always> .

Given that f'(x) > for all x, then it is extremely hard to get f'(x) = so there is no range c these types of is f'(c)=. Rolle’s theorem is ending up failing right here, which is difficult. So we have the Reverse.

The Opposite comes about simply because we built a mistaken guess in the initial location. Remember we guessed that there are at the very least two real roots. So due to the fact the assumption offers us to some thing mathematically illogical. Then we can say that it is unachievable to have at the very least two real roots. It suggests there can only be just a single genuine root. Thus, we answered this problem.

Permit us search at the other Rolle’s Theorem instance.

Dilemma No. 6: Let f(x) = x^2 + 4x -5f(x)=x
2
+4x−5. Does Rolle’s Theorem guarantees the actuality of c from the interval of [-5,1]?

1st, enable us see if f(x) satisfies the three circumstances.

Is the perform constant from the split of [-5,1]? Properly, the method is polynomial. So it is constant everywhere. So it must continue from the closed interval of [-5,1].

Is the function various from the interval (-5,1)? Perfectly, once more, the procedure is a polynomial. So it must be differentiable everywhere. Hence, it must be further from the open up interval of (-5,1).

Does f(-5)=f(1). Perfectly, we see that:

Equation 6: Rolle’s Theorem example aspect.1

Equation 6: Rolle’s Theorem instance component.1

And

Equation 6: Rolle’s Theorem illustration part.2

Equation 6: Rolle’s Theorem instance portion.1

That’s why we can conclusion that intervals of(-5)=f(1). Because all 3 problems were satisfying. Then Rolle’s Theorem guarantees the existence of c. To locate c, we resolve for f'(x)= and check irrespective of whether -5 < x < 1. Noticing that

Equation 6: Rolle’s Theorem example part.3

Equation 6: Rolle’s Theorem example part.3

Setting it equal to 0 gives

Equation 6: Rolle’s Theorem example part.4

Equation 6: Rolle’s Theorem example part.4

Different for x gives:

Equation 6: Rolle’s Theorem example part.5

Equation 6: Rolle’s Theorem example part.5

Notice if -5 < x = -2 < 1, so we must say the number c = -2. If one wants to do more practice on these questions, we suggest that one click the link down.

http://tutorial.math.lamar.edu/Problems/CalcI/MeanValueTheorem.aspx

Here is a lot of practice on these questions involving Rolle’s Theorem. There is also the step by step solutions.

Now that we know with Rolle’s Theorem let us take a look at the Mean Value Theorem.

What is the Mean Value Theorem? In differentiable calculations.

The Mean Value Theorem is the expansion of Rolle’s Theorem. The Mean Value Theorem means that there exists a number c such that a < c < b, and

Formula No.1: Mean Value Theorem

Formula No. 1: Mean Value Theorem

Even though the word means is in this theorem. One can see that it has nothing to do with average. The equation does not even relate to the formula for a mean. Anyway, let us look at how we can prove this theorem.

Mean Value Theorem Proof

To proving Mean Value Theorem, we need to make these guesses again:
Let f(x) satisfy the following conditions:

f(x) is continuous on the interval of [a,b]

f(x) is differentiable on the interval of (a,b)

Now, find two points (a, f(a)) and (b, f(b)) in the function. Then it is going to draw a line between those points. Call that the secant equation.

Graph 1: Secant line

Graph 1: Secant line

We see this is a straight line. So from this graph, one can end that we need to write an equation for this straight line. Recalling that the equation of a line in point-slope form is:

Equation 7: Mean Value Theorem Proof part.1

Equation 7: Mean Value Theorem Proof part.1

The equation of the Secant line is:

Equation 7: Mean Value Theorem Proof part.2

Equation 7: Mean Value Theorem Proof part.2

Moves to f(a) to another side of the equation will give:

Equation 7: Mean Value Theorem Proof part.3

Equation 7: Mean Value Theorem Proof part.3

Now we will be going to create another function is called g(x). The process is a subtraction between the function f(x) and the secant line y. In other words,

Equation 7: Mean Value Theorem Proof part.4

Equation 7: Mean Value Theorem Proof part.4

Notice of g(x) is also continuous and different because subtraction of two linear and

differentiated functions is still continuous and differentiable. Hence, we can take the derivative. Taking the result that will give us:

Equation 7: Mean Value Theorem Proof part.5

Equation 7: Mean Value Theorem Proof part.5

Now we have to relate this to Rolle’s Theorem somehow. Rolle’s Theorem is holding for three conditions, but we only have two. In other words, we should know that g(x) is continuing and different. But we were missing the condition where g(a) = g(b). Lets us see if this is true.

Equation 7: Mean Value Theorem Proof part.6

Equation 7: Mean Value Theorem Proof part.6

Notice if g(b) =g(a), and therefore g(x) satisfies the three conditions of Rolle’s

Theorem. Thus, we know that there is a number c such that g'(c) = 0 within the interval of [a,b]. We can also use this fact for g'(x). Recall g'(x) was

Equation 7: Mean Value Theorem Proof part.7

Equation 7: Mean Value Theorem Proof part.7

So if g'(c) = 0, then

Equation 7: Mean Value Theorem Proof part.8

Equation 7: Mean Value Theorem Proof part.8
It would indicate that

Equation 7: Mean Value Theorem Proof part.9

Equation 7: Mean Value Theorem Proof part.9

So we just proved the Mean Value Theorem if one is interested in the Mean Value Theorem. For derivative and Mean Value Theorem for integrals. Then we suggest one look at the Mean Value Theorem section.

So, this is vital information on the topic of Roll’s Theorem. I have mentioned here the simple steps of Roll’s Theorem. With the Formulas. Make your sum simple in applying the helpful steps with formulas in Roll’s Theorem.

If Queries or Questions is persisting then, please feel free to comment on the viewpoints.

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