on the interval(*a, b*)and is continuous on the interval[*a, b*]. Then

*f*′(*c*)=*f*(*b*)−*f*(*a*)

*b*_{−}*a*

for some*a < c < b*.

*a* *c* _{b}*x*

*y*

Figure 10.11: A geometric interpretation of the Mean Value Theorem

**Proof***Let*

*m*=*f*(*b*)−*f*(*a*)

*b*_{−}*a* *,*

*and consider a new function g*(*x*) = *f*(*x*)_{−}*m*(*x*_{−}*a*)_{−}*f*(*a*)*. We know that*
*g*(*x*) *has a derivative on* [*a, b*]*, since g*′(*x*) = *f*′(*x*)_{−}*m. We can compute*
*g*(*a*)=*f*(*a*)_{−}*m*(*a*−*a*)_{−}*f*(*a*)=0*and*

*g*(*b*)=*f*(*b*)_{−}*m*(*b*_{−}*a*)_{−}*f*(*a*)=*f*(*b*)_{−}*f*(*b*)−*f*(*a*)

*b*−*a* (*b*−*a*)−*f*(*a*)

=*f*(*b*)_{−}(*f*(*b*)_{−}*f*(*a*))_{−}*f*(*a*)
=0*.*

*So g*(*a*)=*g*(*b*)=0*. Now by Rolle’s Theorem, that at some c,*
*g*′_{(}_{c}_{)}_{=}_{0} _{for some a < c < b.}

*But we know that g*′_{(}_{c}_{)}_{=}* _{f}*′

_{(}

_{c}_{)}

_{−}

_{m, so}0=*f*′(*c*)_{−}*m*=*f*′(*c*)_{−}*f*(*b*)−*f*(*a*)

*Hence*

*f*′_{(}_{c}_{)}_{=}*f*(*b*)−*f*(*a*)

*b*_{−}*a* *.*

We can now answer our second question above.

**Example 10.3.4** Suppose you drive a car from toll booth on a toll road to
another toll booth 30 miles away in half of an hour. Must you have been
driving at 60 miles per hour at some point?

**Solution***If p*(*t*)*is the position of the car at time t, and*0*hours is the starting*

*time with*1*/*2*hours being the ﬁnal time, the Mean Value Theorem states there*
*is a time c*

*p*′(*c*)=30−0

1*/*2 =60 *where*0*< c <*1*/*2*.*

*Since the derivative of position is velocity, this says that the car must have been*
*driving at*60*miles per hour at some point.*

Now we will address the unthinkable, could there be a function *f*(*x*) whose
derivative is zero on an interval that is not constant? As we will see, the answer is
“no.”

**Theorem 10.3.5** If*f*′_{(}_{x}_{)}_{=}_{0}_{for all}_{x}_{in an interval}_{I}_{, then}_{f}_{(}_{x}_{)}_{is constant on}

*I*.

**Proof***Let a < b be two points in I. By the Mean Value Theorem we know*

*f*(*b*)_{−}*f*(*a*)

*b*_{−}*a* =*f*

′_{(}_{c}_{)}

*for some c in the interval*(*a, b*)*. Since f*′(*c*)=0*we see that f*(*b*)=*f*(*a*)*. Moreover,*
*since a and b were arbitrarily chosen, f*(*x*)*must be the constant function.*

**Example 10.3.6** Suppose two diﬀerent functions have the same derivative.
What can you say about the relationship between the two functions?

**Solution***Set h*(*x*)=*f*(*x*)_{−}*g*(*x*)*, so h*′_{(}_{x}_{)}_{=}* _{f}*′

_{(}

_{x}_{)}

_{−}

*′*

_{g}_{(}

_{x}_{)}

*′*

_{. Now h}_{(}

_{x}_{)}

_{=}

_{0}

_{on the}*interval*(*a, b*)*. This means that h*(*x*)=*k where k is some constant. Hence*
*g*(*x*)=*f*(*x*)+*k.*

**Example 10.3.7** Describe all functions whose derivative issin(*x*).

**Solution***One such function is* −cos(*x*)*, so all such functions have the form*

−cos(*x*)+*k, see Figure10.12.*

1 2 3 4 5 6
−4
−2
2
4
*x*
*y*

Figure 10.12: Functions of the form−cos(*x*)+*k*,
each of whose derivative issin(*x*).

**Exercises for Section 10.3**

(1) Let*f*(x)_{=}*x*2. Find a value*c*_{∈}(−1*,*2)so that*f*′_{(c)}_{equals the slope between the endpoints}
of*f*(x)on[−1*,*2].

### ➠

(2) Verify that*f*(x)=*x/(x*+2)satisﬁes the hypotheses of the Mean Value Theorem on the
interval[1*,*4]and then ﬁnd all of the values,*c*, that satisfy the conclusion of the theorem.

### ➠

(3) Verify that*f*(x) =3*x/(x*+7)satisﬁes the hypotheses of the Mean Value Theorem on
the interval[−2*,*6]and then ﬁnd all of the values,*c*, that satisfy the conclusion of the
theorem.

### ➠

(4) Let*f*(x)_{=}tan(x). Show that*f*(π)_{=}*f*(2*π)*_{=}0 but there is no number*c*_{∈}(π,2*π)*such
that*f*′_{(c)}_{=}_{0. Why does this not contradict Rolle’s theorem?}

_{➠}

(5) Let*f*(x) _{=}(x_{−}3)−2. Show that there is no value*c*_{∈} (1*,*4)such that*f*′_{(c)} _{=}_{(f}_{(}_{4}_{)}_{−}

*f*(1))/(4−1). Why is this not a contradiction of the Mean Value Theorem?

### ➠

(6) Describe all functions with derivative*x*2_{+}47*x*_{−}5.

### ➠

(7) Describe all functions with derivative 1

1+*x*2.

### ➠

(8) Describe all functions with derivative*x*3−1

*x*.

### ➠

(9) Describe all functions with derivativesin(2*x)*.

### ➠

(10) Show that the equation 6*x*4−7*x*+1=0 does not have more than two distinct real roots.

### ➠

(11) Let*f*(x)be diﬀerentiable onR_{. Suppose that}*f*′_{(x)},0 for every*x*. Prove that*f* has at
most one real root.

### ➠

**11.1**

**Basic Antiderivatives**

Computing derivatives is not too diﬃcult. At this point, you should be able to take the derivative of almost any function you can write down. However, undoing derivatives is much harder. This process of undoing a derivative is called taking an

*antiderivative*.

**Definition** A function*F*(*x*)is called an**antiderivative**of*f*(*x*)on an interval if

*F*′(*x*)=*f*(*x*)
for all*x* in the interval.

We have special notation for the antiderivative:

**Definition** The antiderivative is denoted by
Z

*f*(*x*)*dx*=*F*(*x*)+*C,*

where*dx*identiﬁes*x*as the variable and*C*is a constant indicating that there
are many possible antiderivatives, each varying by the addition of a constant.
This is often called the**indefinite integral**.

Here are the basic antiderivatives. Note each of these examples comes directly from our knowledge of basic derivatives.

**Theorem 11.1.1 (Basic Antiderivatives)**
•
Z
*k dx*=*kx*+*C*.
•
Z
*xndx*= *x*
*n*+1
*n*+1 +*C* (*n*,−1).
•
Z
*exdx* =*ex*+*C*.
•
Z
*axdx*= *a*
*x*
ln(*a*)+*C*.
•
Z _{1}
*x* *dx*=ln|*x*|+*C*.
•
Z
cos(*x*)*dx* =sin(*x*)+*C*.
•
Z
sin(*x*)*dx*=−cos(*x*)+*C*.
•
Z
tan(*x*)*dx*=−ln_{|}cos(*x*)_{|}+*C*.
•
Z
sec2(*x*)*dx*=tan(*x*)+*C*.
•
Z
csc2(*x*)*dx*=−cot(*x*)+*C*.
•
Z

sec(*x*) tan(*x*)*dx*=sec(*x*)+*C*.

•
Z
csc(*x*) cot(*x*)*dx*=−csc(*x*)+*C*.
•
Z _{1}
*x*2_{+}_{1}*dx*=arctan(*x*)+*C*.
•
Z _{1}
√
1−*x*2*dx*=arcsin(*x*)+*C*.

It may seem that one could simply memorize these antiderivatives and antidiﬀeren-
tiating would be as easy as diﬀerentiating. This is**not**the case. The issue comes
up when trying to combine these functions. When taking derivatives we have the

*product rule*and the*chain rule*. The analogues of these two rules are much more
diﬃcult to deal with when taking antiderivatives. However, not all is lost. We have
the following analogue of the Sum Rule for derivatives, Theorem3.2.6.

**Theorem 11.1.2 (The Sum Rule for Antiderivatives)** Given two functions

*f*(*x*)and*g*(*x*)where*k* is a constant:

•
Z
*kf*(*x*)*dx*=*kF*(*x*)+*C*.
•
Z
(*f*(*x*)+*g*(*x*))*dx*=*F*(*x*)+*G*(*x*)+*C*.

Let’s put this rule and our knowledge of basic derivatives to work.
**Example 11.1.3** Compute

Z