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# Find an inverse for the function log |x|

Consider the function for . Prove that this function has an inverse, determine the domain of this inverse, and find a formula to compute the inverse .

Proof. From the discussion on page 146 of Apostol we know that a function which is continuous and strictly monotonic on an interval has an inverse on . The function is continuous and strictly monotonic on the negative real axis; therefore, it has an inverse. We know it is continuous since the log function is continuous on the positive real axis, and for all , in particular, for all . Furthermore, we know it is strictly monotonic since

Therefore, has an inverse for all . The domain of this inverse is the range of which is all of

To find a formula for the inverse we set

Therefore, valid for all .

A sketch for the graph of is given by

# Prove that 2 cosh2 (x/2) = cosh x + 1

Prove the following identity,

Proof. Computing directly from the definition of the hyperbolic cosine,

# Prove that 2 sinh2 (x/2) = cosh x – 1

Prove the following identity:

Proof. We compute directly from the definition of ,

# Prove that (cosh x + sinh x)n = cosh (nx) + sinh (nx)

Prove the following identity:

Proof. We know from a previous exercise (Section 6.19, Exercise #9) that

Therefore, we have

# Prove that cosh x – sinh x = e-x

Prove that

Proof. Using the definition of the hyperbolic functions in terms of the exponential, we have

# Prove that cosh x + sinh x = ex

Prove that

Proof. We use the definitions of hyperoblic sine and cosine to compute,

# Prove formulas for the partial derivatives of xy

Define a function of two variables, with ,

Prove

Proof. First, we write,

Then we compute the derivatives using the chain rule and formulas for the derivatives of the exponential and logarithm,

And,

# Prove some inequalities of the exponential function

For and for prove the following inequalities,

where we assume in the second inequality. Use this to show

Proof. From the previous exercise (Section 6.17, Exercise #41) we know

for . But by assumption we have and ; hence, . Therefore this inequality must hold for :

Again, by the previous exercise we have

for . Since and this implies . Therefore,

Letting , we have

# Prove some inequalities using the function ex-1-x

1. Define a function:

Prove that for all and for all . Prove the following inequalities are valid for all ,

2. Using integration and part (a) prove

3. Using integration and part (a) prove

4. State and prove a generalization of the inequalities above.

1. First, we take the derivative of ,

Since is strictly increasing everywhere (since for all ) we have for and for . Thus, has a minimum at and so for all . Therefore, when ; hence,

Furthermore, when we have

Therefore, if then and so we have

2. Using the inequalities in part (a) we integrate over the interval to ,

For the other inequality we proceed similarly,

3. We use the same strategy as before, starting with the inequalities we established in part (b),

Similarly, for the other inequality,

4. Claim: The following general inequalities hold for all :

Proof. The proof is by induction. We have already established the inequalities for the case for all three inequalities. Now, to prove the first inequality assume

Then we have

This establishes the first inequality. For the second inequality we have proved the case already. Assume

Since is odd we may write for some nonnegative integer . We want to show that the inequality must hold for the next odd integer, . We have,

We want to show that the inequality holds for so we integrate both sides again,

Hence, if the inequality holds for odd , then it also holds for the next odd number, . Hence, it holds for all positive, odd integers.
The exact same induction argument works for all of the even integers (starting with the case we proved in part (c))

# Prove properties of a function satisfying a given functional equation

Assume is a function defined for all such that

For parts (b)-(d) assume also that is differentiable for all real (i.e., assume exists everywhere).

1. Using the given equation above, prove or . Prove that if then for all .
2. Prove that for all .
3. Prove that for some constant .
4. Prove that if then .

1. Proof. Assume . Then if by the functional equation we have

Therefore, if then we must have . Hence, we have either or .

Furthermore, if , then for any by the functional equation we have

Hence, for any

2. Proof. Since the derivative of exists everywhere by hypothesis, we use the limit definition of the derivative and the functional equation to write,

3. Proof. First, if then by the functional equation we know for any we have

So, if then is the constant function which is 0 everywhere. Hence, for all .
Next, we consider the case that . Since exists for all we use the definition of derivative and the functional equation again to write,

The final line follows since exists by assumption so we can write for some

4. Proof. Since (by part (c)) we know, by the previous exercise (Section 6.17, Exercise #39) that for some constant . Since by assumption, we know by part (a) that . Hence,