# Calculus 1 : Local Minimum

## Example Questions

### Example Question #1491 : Functions

Find the local minima for the following function on the interval :

Explanation:

To find the local minima of a function, we must find the point(s) at which the first derivative changes from negative to positive.

The first derivative of the function is:

and was found using the following rule:

Now, we must find the critical points, the points at which the first derivative is zero:

Now, we create intervals over which to see whether the first derivative is positive or negative:

On the first interval, the first derivative is negative, while on the second interval it is positive, indicating that the point of the change in sign  is the local minimum.

### Example Question #11 : Local Minimum

What is the minimum of over the interval ?

Explanation:

To find the minimum of a function, find the first derivative. In order to find the derivative of this fuction use the power rule which states, .

Given the function,  and applying the power rule we find the following derivative.

Check the -value at each endpoint and when the first derivative is zero, namely

The smallest value is .

### Example Question #11 : How To Find Local Minimum Graphing Functions Of Curves

Find the -value where the local minimum occurs on

.

Explanation:

To find the minimum of a function, find the first derivative. In order to find the derivative of this fuction use the quotient rule which states,

.

Given the function,  and applying the quotient rule we find the following derivative.

when  and  when , which indicates that has a local minimum at .

### Example Question #42 : Graphing Functions

Find the relative minimum of .

x

Explanation:

To find the relative minimum, you must first find the derivative of the function so that you can find the critical points. When taking the derivative, multiply the exponent by the leading coefficient and then subtract 1 from the exponent: . Then, set that equal to 0 to get your critical points: . Then, test on either side of that point to see what the behavior of the function is. To the left of 2 (plugging in 0, for example), the value is negative. To the right, the value is positive. Since the slope of the function is going from negative to positive at x=2, there is a relative minimum at x=2.

### Example Question #51 : Graphing Functions

Find the local minima of the following function:

Explanation:

To find the local minima of the function, we must find the x value at which the function's first derivative changes from negative to positive.

First, we must find the first derivative of the function:

The derivative was found using the following rules:

Now, we must find the critical values (where the first derivative is equal to zero) by setting the first derivative equal to zero:

Next, we must create the intervals in which the critical value is the upper and lower limit, respectively:

On the first interval, the first derivative is less than one, and on the second interval, the first derivative is positive (simply plug in any point on the interval into the first derivative function and check the sign). The change from negative to positive occurs at x=3, so  is our answer.

### Example Question #11 : Local Minimum

Where is the local minimum of

Explanation:

To find the local minimum, first find the derivative of the function. To take the derivative, multiply the exponent by the coefficient in front of the x term and then decrease the exponent by 1. Therefore, the derivative is: . Then, set that equal to 0 to get your critical point(s): . Put that point on a number line and test values on either side. To the left of 2, the values are negative. To the right of 2, the values are positive. Therefore, the slope goes from negative to positive, indicating there is a local min at x=2.

### Example Question #53 : Graphing Functions

Explanation:

The first step is to chop up the expression into two terms: . Now, integrate each term. Whenever there is an x on the denominator, the integral is lnx. Multiply that by the numerator of 4. The integral of 1 is x. Put those together to get your answer of .

### Example Question #1501 : Functions

Find where the local minima occur for the following function:

The function is always increasing

Explanation:

To find where the local minima occur for the function, we need the x-values at which the first derivative changes from negative to positive.

The first derivative is

and was found using the following rule:

Now, we must find the critical values at which the first derivative is equal to zero:

Now, using the critical values, we make the intervals on which we see whether the first derivative is positive or negative (plug in any value on the interval into the first derivative function):

On the first interval, the first derivative is positive, on the second it is negative, and on the third it is positive. So, a local minimum occurs at

because the first derivative changed from positive to negative at that point.

### Example Question #54 : Graphing Functions

Find the x value of the local minima of .

Explanation:

We need to differentiate term by term, applying the power rule,

This gives us

The critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula:

Any local minimum will fall at a critical point where the derivative passes from negative to positive. To check this, we check a point in each of the intervals defined by the critical points:

.

Let's take -2 from the first interval, 0 from the second interval, and 2 from the third interval.

The derivative moves from negative to positive at 1, so that is the function's only local minimum.

### Example Question #55 : Graphing Functions

Find the local minima of the following function:

Explanation:

To determine the local minima for the function, we must determine where the first derivative changes from negative to positive.

The first derivative of the function is equal to

and was found using the following rules:

Now, we must find the critical values, the values at which the first derivative equals zero:

We use the critical value to make our intervals on which we check the sign of the first derivative:

Note that at the endpoints of the intervals, the first derivative is neither positive nor negative.

Next, we plug in any value on the intervals into the first derivative and check the sign. On the first interval, the first derivative is negative, while on the second, it is positive. Thus, a local mininum exists at .