### Topics

## Introduction to Second Derivatives And Beyond - At A Glance:

### Sample Problem

- Below is a graph of a function
*f* with a minimum at *x* = x_{0}. Determine the sign of the derivative *f*' at each labelled *x*-value.

- Below is a graph of a function
*f* with a maximum at *x* = x_{0}. Determine the sign of the derivative *f *' at each labelled *x*-value.

A minimum, assuming it's not at the endpoint of an interval, usually looks like this:

The derivative is zero (or undefined) at the place the minimum occurs:

Since the function must decrease down to the minimum and then increase away from the minimum, the derivative is negative to the left and positive to the right of the place where the minimum occurs:

We can use a numberline to keep track of the sign of *f *' like this:

A maximum, assuming it's not at the endpoint of an interval, usually looks like this:

The derivative is zero (or undefined) at the place the maximum occurs:

Since the function must increase up to the maximum and then decrease away from the maximum, the derivative is positive to the left and negative to the right of the place where the maximum occurs:

We can use a numberline to keep track of the sign of *f *' like this:

If we don't have a graph of the function, we can go the other way around: we make a numberline first, and use that to determine if a critical point of *f* is a maximum or a minimum or neither. We find the sign of *f *' a little to the left of the critical point, and a little to the right of the critical point. If we find this, the critical point is a minimum:

If we find this, the critical point is a maximum:

If we find one of these, the critical point is neither a min nor a max:

This process is called the **First Derivative Test** because we are using the first derivative to test whether a critical point is a min or a max or neither.

#### Example 1

Let *f* (*x*) = *xe*^{x}. Use the First Derivative Test to determine if each critical point is a minimum, a maximum, or neither. | |

First we need to find the critical point(s). The derivative is *f *'(*x*) = *xe*^{x} + *e*^{x} = *e*^{x}(*x* + 1).
This is zero only when *x* = -1. Since *e*^{x} is always positive, the sign of *f*'(*x*) is the same as the sign of ( *x* + 1 ). When *x* < -1, the quantity ( *x* + 1 ) is negative, so *f*'(*x*) is also. When *x* > -1 the quantity ( *x* + 1 ) is positive, so *f*'(*x*) is also. We therefore find this numberline:
Therefore *f* (*x*) is decreasing until *x* approaches -1, then increasing again, so at *x* = -1 there is a minimum. | |

#### Example 2

Let *f* (*x*) = *x*^{3}. Use the First Derivative Test to determine if each critical point is a minimum, a maximum, or neither.
| |

The derivative is *f *'(*x*) = 3*x*^{2},
which is never undefined and is zero only when *x* = 0. Since *x*^{2} is positive when *x* ≠ 0, *f *'(*x*) is also positive when *x* ≠ 0. We find this numberline: Since *f* (*x*) is increasing both to the left and to the right of *x* = 0, *x* = 0 is neither a minimum nor a maximum. | |

#### Example 3

Let *f* (*x*) = sin *x* on the interval 0 ≤ *x* ≤ 2π. Use the First Derivative Test to determine if each critical point is a minimum, a maximum, or neither | |

We take the derivative: *f *'(*x*) = cos *x*.
The derivative is never undefined and is zero when and when (remember, we're only looking at the interval [0,2π] right now). So far, we have a numberline that looks like this: In order to use the first derivative test, we need to know what the sign of *f *' is everywhere on the numberline. This is possible to see from a graph of *f *'(*x*): The numberline now looks like this: Since *f*' is positive to the left and negative to the right of , the function *f* has a maximum at . Since *f *' is negative to the left and positive to the right of , the function *f* has a minimum at . | |

#### Exercise 1

For the function, use the First Derivative Test to determine if each critical point is a minimum, a maximum, or neither.

for 0 < x

Answer

We use the quotient rule to find the derivative:

This is defined for all *x* in the domain of *f*, that is, all *x* greater than zero. It's zero when

*ln* *x* = 1

which means when

*x* = *e*.

Since the denominator of the derivative is (*ln* *x*)^{2}, which is always positive, the sign of *f*'(*x*) is determined by the sign of the numerator

*ln* *x* - 1.

This is negative when *x* < *e* and positive when *x* > *e*, so we find this numberline:

Therefore *f* (*x*) is decreasing to *x* = *e* and then increasing, so *f* (*x*) hits a minimum at *x* = *e*.

#### Exercise 2

For the function, use the First Derivative Test to determine if each critical point is a minimum, a maximum, or neither.

*f* (*x*) = -*x*^{3} + 3*x*^{2} - 3*x*

Answer

The derivative is

This is only zero when *x* = 1, and never undefined, so *x* = 1 is the only critical point. Since the quantity ( *x* - 1 )^{2} is positive for all *x* ≠ 1, the derivative

*f*'(*x*) = -3( *x* - 1 )^{2}

is negative for all *x* ≠ 1. We find this numberline:

Since *f* is decreasing, levels out, and then decreases again, we have neither a minimum nor a maximum at *x* = 1.

#### Exercise 3

For the function, use the First Derivative Test to determine if each critical point is a minimum, a maximum, or neither.

Answer

We use the quotient rule to find the derivative:

*f*'(*x*) is undefined only at *x* = -3, in which case *f* is also undefined so this is not a critical point. *f*'(*x*) is zero when

*x*^{2} + 6*x* + 5 = 0.

Happily, this quadratic factors as

*x*^{2} + 6*x* + 5 = (*x* + 5)(*x* + 1),

so *f*'(*x*) is zero at *x* = -5 and *x* = -1. Here's the numberline so far:

To figure out whether *f* has a maximum or a minimum at each of these critical points, we need to find the sign of the first derivative so we can fill in the numberline.

The derivative

*f*'(*x*) = (*x* + 5)(*x* + 1)

is negative when exactly one of the quantities (*x* + 5) or (*x* + 1) is negative. This is only possible when

-5 < *x* < -1,in which case (*x* + 1) is negative but (*x* + 5) is positive. So the numberline looks like this:

Since *f*' is positive to the left and negative to the right of *x* = -5, the function *f* has a maximum at *x* = -5. Since *f*' is negative to the left and positive to the right of *x* = -1, the function *f* has a minimum at *x* = -1.

#### Exercise 4

For the function, use the First Derivative Test to determine if each critical point is a minimum, a maximum, or neither.

Answer

We use the quotient rule to find the derivative:

This is only undefined when *x*^{2}-3*x* + 2 = 0, but such an *x* value doesn't count as a critical point since it also makes *f* undefined. The derivative is zero when the numerator is zero, which occurs when

Now we have this numberline:

Since the denominator of *f*' is always positive, the sign of *f*' is the same as the sign of its numerator. The numerator

-6*x*^{2} + 12

is positive when 6*x*^{2} is smaller than 12, that is, when *x*^{2} is smaller than 2. When *x* is within sqrt 2 of 0 (that is, between -sqrt 2 and + sqrt 2),

-6*x*^{2} + 12 > 0.

Otherwise,

-6*x*^{2} + 12 < 0.

On the numberline, we have this:

Since *f*' is negative to the left of *x* = -sqrt 2 and positive to the right of *x* = -sqrt 2, *f* has a minimum at *x* = -sqrt2. Since *f*' is positive to left of *x* = sqrt 2 and negative to the right, *f* has a maximum at *x* = sqrt 2.

#### Exercise 5

For the function, use the First Derivative Test to determine if each critical point is a minimum, a maximum, or neither.

*f* (*x*) = *e*^{x}

Answer

The derivative of *f* (*x*) is

*f*'(*x*) = *e*^{x}.

Since *f*'(*x*) is always positive, the function *f* (*x*) has no critical points and therefore cannot have any maxima or minima.