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# Use derivatives to sketch the graph of a function

Let

1. Find all points such that ;
2. Determine the intervals on which is monotonic by examining the sign of ;
3. Determine the intervals on which is monotonic by examining the sign of ;
4. Sketch the graph of .

1. We take the derivative,

Thus,

2. is increasing for and , and decreasing for .
3. Taking the second derivative,

Thus, is increasing if , and decreasing if .

4. We sketch the curve,

# Use derivatives to sketch the graph of a function

Let

1. Find all points such that ;
2. Determine the intervals on which is monotonic by examining the sign of ;
3. Determine the intervals on which is monotonic by examining the sign of ;
4. Sketch the graph of .

1. We take the derivative,

Thus,

2. is increasing for , and decreasing for .
3. Taking the second derivative,

Thus, is increasing if , and decreasing if .

4. We sketch the curve,

# Use derivatives to sketch the graph of a function

Let

1. Find all points such that ;
2. Determine the intervals on which is monotonic by examining the sign of ;
3. Determine the intervals on which is monotonic by examining the sign of ;
4. Sketch the graph of .

1. We take the derivative,

Thus,

2. Since when and when we have is increasing if and is decreasing if .
3. Taking the second derivative,

Thus, is increasing for ; and decreasing for .

4. We sketch the curve,

# Use derivatives to sketch the graph of a function

Let

1. Find all points such that ;
2. Determine the intervals on which is monotonic by examining the sign of ;
3. Determine the intervals on which is monotonic by examining the sign of ;
4. Sketch the graph of .

1. We take the derivative,

Thus,

2. Since when and when we have is decreasing if and is increasing if .
3. Taking the second derivative,

Since this is positive for all , this means is increasing for all .

4. We sketch the curve,

# Prove the second derivative of a function with a given property must have a zero

Consider a function which is continuous everywhere on an interval and has a second derivative everywhere on the open interval . Assume the chord joining two points and on the graph of the function intersects the graph of the function at a point with . Prove there exists a point such that .

Proof. Let be the equation of the line joining and . Then define a function

Since and intersect at the values , and this means

By our definition of then we have

Further, since and are continuous and differentiable on , we apply Rolle’s theorem twice: first on the interval and then on the interval . These two applications of Rolle’s theorem tells us there exist points and such that

Then, we apply Rolle’s theorem for a third time, this time to the function on the interval to conclude that there exists a such that . Then, since we know (since ). So,

Thus, for some

# Prove properties about the zeros of a polynomial and its derivatives

Consider a polynomial . We say a number is a zero of multiplicity if

where .

1. Prove that if the polynomial has zeros in , then its derivative has at least zeros in . More generally, prove that the th derivative, has at least zeros in the interval.
2. Assume the th derivative has exactly zeros in the interval . What can we say about the number of zeros of in the interval?

1. Proof. Let denote the distinct zeros of in and their multiplicities, respectively. Thus, the total number of zeros is given by,

By the definition given in the problem, if is a zero of of multiplicity then

Taking the derivative (using the product rule), we have

Thus, again using the definition given in the problem, is a zero of of multiplicity .
Next, we know from the mean-value theorem for derivatives, that for distinct zeros and of there exists a number (assuming, without loss of generality, that ) such that . Hence, if has distinct zeros, then the mean value theorem guarantees numbers such that . Thus, has at least:

By induction then, the th derivative has at least zeros.

2. If the th derivative has exactly zeros in , then we can conclude that has at most zeros in .

# Show that x^2 = x*sin x + cos x for exactly two real numbers x

Consider the equation

Show that there are two values of such that the equation is satisfied.

Proof. Let . (We want then to find the zeros of this function since these will be the points that .) Then,

Since for any (since ), we have

Then, is continuous and differentiable everywhere, so we may apply Rolle’s theorem on any interval. So, by Rolle’s theorem we know has at most two zeros (if there were three or more, say and , then there must be distinct numbers and with such that , but we know there is only one such that ).
Furthermore, has at least two zeros since , , and . Thus, by Bolzano’s theorem there are zeros between each of these points. We have that the number of zeros of is at most two and at least 2. Hence, the number of zeros must be exactly two

# Explain why the absence of a zero does not violate Rolle’s theorem

Consider the function

Show that and , but that the derivative for all . Explain why this does not violate Rolle’s theorem.

Proof. First, we show that and by a direct computation:

Then, we compute the derivative,

To show for any we consider three cases:

• If then implies (since times a negative is positive).
• If then implies (since times a positive is then negative).
• If , then is undefined (since ).

Thus, for any

This is not a violation of Rolle’s theorem since the theorem requires that be differentiable for all on the open interval . Since is not defined at , we have is not differentiable on the whole interval. Hence, Rolle’s theorem does not apply.

# Prove there is only one zero of x3 – 3x + b in [-1,1]

Prove, using Rolle’s theorem, that for any value of there is at most one such that

Proof. The argument is by contradiction. Suppose there are two or more points in for which . Let be two such points with . Since are both in we have

Since is continuous on and differentiable on (all polynomials are continuous and differentiable everywhere), we may apply Rolle’s theorem on the interval to conclude that there is some such that

But this contradicts that (since and ). Hence, there can be at most one point such that

# Prove there is exactly one negative solution to an equation

Show there is exactly one such that for and an odd, positive integer.

Proof. Let , and let with . Then, (for odd ) so . Since and , by the Intermediate Value Theorem, we know takes every value between and 0 for some . Thus, we know there exists such that (since ). This implies for some .

We know this solution is unique since is strictly increasing on the whole real line for odd