Relative Minimum - AP Calculus AB

We notice that this function has 2 extrema which are located at x=-1 and x=0. We could have also of found this by looking at the function itself:

.

We know that extrema exist when the slope of the function is zero, hence we take the derivative, set it equal to zero, and than solve for x.

The derivative is the following:

.

Therefore setting both pieces equal to zero we see the following:

and .

Finding out if one of these x values produce a local min or max however requires either the first derivative test, the second derivative test, or analyzing the graph. We see that for a small neighborhood around , is the largest term, hence it is a local max. Similarly, for a small neighborhood around , is the smallest term, hence is a local min and the only local min of this function. Remember that it is not the x value that must be the smallest, rather it is it's corresponding y value.

The mean value theorem states that if continuous has , then for all , there must be an that maps to , so since the value above go from negative to positive twice, zero must be mapped to at least twice.

We notice that this function has 2 extrema which are located at x=-1 and x=0. We could have also of found this by looking at the function itself:

.

We know that extrema exist when the slope of the function is zero, hence we take the derivative, set it equal to zero, and than solve for x.

The derivative is the following:

.

Therefore setting both pieces equal to zero we see the following:

and .

Finding out if one of these x values produce a local min or max however requires either the first derivative test, the second derivative test, or analyzing the graph. We see that for a small neighborhood around , is the largest term, hence it is a local max. Similarly, for a small neighborhood around , is the smallest term, hence is a local min and the only local min of this function. Remember that it is not the x value that must be the smallest, rather it is it's corresponding y value.

The mean value theorem states that if continuous has , then for all , there must be an that maps to , so since the value above go from negative to positive twice, zero must be mapped to at least twice.

The mean value theorem states that if continuous has , then for all , there must be an that maps to , so since the value above go from negative to positive twice, zero must be mapped to at least twice.

We notice that this function has 2 extrema which are located at x=-1 and x=0. We could have also of found this by looking at the function itself:

.

We know that extrema exist when the slope of the function is zero, hence we take the derivative, set it equal to zero, and than solve for x.

The derivative is the following:

.

Therefore setting both pieces equal to zero we see the following:

and .

Finding out if one of these x values produce a local min or max however requires either the first derivative test, the second derivative test, or analyzing the graph. We see that for a small neighborhood around , is the largest term, hence it is a local max. Similarly, for a small neighborhood around , is the smallest term, hence is a local min and the only local min of this function. Remember that it is not the x value that must be the smallest, rather it is it's corresponding y value.

The mean value theorem states that if continuous has , then for all , there must be an that maps to , so since the value above go from negative to positive twice, zero must be mapped to at least twice.

We notice that this function has 2 extrema which are located at x=-1 and x=0. We could have also of found this by looking at the function itself:

.

We know that extrema exist when the slope of the function is zero, hence we take the derivative, set it equal to zero, and than solve for x.

The derivative is the following:

.

Therefore setting both pieces equal to zero we see the following:

and .

Finding out if one of these x values produce a local min or max however requires either the first derivative test, the second derivative test, or analyzing the graph. We see that for a small neighborhood around , is the largest term, hence it is a local max. Similarly, for a small neighborhood around , is the smallest term, hence is a local min and the only local min of this function. Remember that it is not the x value that must be the smallest, rather it is it's corresponding y value.

The mean value theorem states that if continuous has , then for all , there must be an that maps to , so since the value above go from negative to positive twice, zero must be mapped to at least twice.

We notice that this function has 2 extrema which are located at x=-1 and x=0. We could have also of found this by looking at the function itself:

.

We know that extrema exist when the slope of the function is zero, hence we take the derivative, set it equal to zero, and than solve for x.

The derivative is the following:

.

Therefore setting both pieces equal to zero we see the following:

and .

Finding out if one of these x values produce a local min or max however requires either the first derivative test, the second derivative test, or analyzing the graph. We see that for a small neighborhood around , is the largest term, hence it is a local max. Similarly, for a small neighborhood around , is the smallest term, hence is a local min and the only local min of this function. Remember that it is not the x value that must be the smallest, rather it is it's corresponding y value.

The mean value theorem states that if continuous has , then for all , there must be an that maps to , so since the value above go from negative to positive twice, zero must be mapped to at least twice.

We notice that this function has 2 extrema which are located at x=-1 and x=0. We could have also of found this by looking at the function itself:

.

We know that extrema exist when the slope of the function is zero, hence we take the derivative, set it equal to zero, and than solve for x.

The derivative is the following:

.

Therefore setting both pieces equal to zero we see the following:

and .

Finding out if one of these x values produce a local min or max however requires either the first derivative test, the second derivative test, or analyzing the graph. We see that for a small neighborhood around , is the largest term, hence it is a local max. Similarly, for a small neighborhood around , is the smallest term, hence is a local min and the only local min of this function. Remember that it is not the x value that must be the smallest, rather it is it's corresponding y value.

The mean value theorem states that if continuous has , then for all , there must be an that maps to , so since the value above go from negative to positive twice, zero must be mapped to at least twice.

We notice that this function has 2 extrema which are located at x=-1 and x=0. We could have also of found this by looking at the function itself:

.

We know that extrema exist when the slope of the function is zero, hence we take the derivative, set it equal to zero, and than solve for x.

The derivative is the following:

.

Therefore setting both pieces equal to zero we see the following:

and .

Finding out if one of these x values produce a local min or max however requires either the first derivative test, the second derivative test, or analyzing the graph. We see that for a small neighborhood around , is the largest term, hence it is a local max. Similarly, for a small neighborhood around , is the smallest term, hence is a local min and the only local min of this function. Remember that it is not the x value that must be the smallest, rather it is it's corresponding y value.

The mean value theorem states that if continuous has , then for all , there must be an that maps to , so since the value above go from negative to positive twice, zero must be mapped to at least twice.

We notice that this function has 2 extrema which are located at x=-1 and x=0. We could have also of found this by looking at the function itself:

.

We know that extrema exist when the slope of the function is zero, hence we take the derivative, set it equal to zero, and than solve for x.

The derivative is the following:

.

Therefore setting both pieces equal to zero we see the following:

and .

Finding out if one of these x values produce a local min or max however requires either the first derivative test, the second derivative test, or analyzing the graph. We see that for a small neighborhood around , is the largest term, hence it is a local max. Similarly, for a small neighborhood around , is the smallest term, hence is a local min and the only local min of this function. Remember that it is not the x value that must be the smallest, rather it is it's corresponding y value.

The mean value theorem states that if continuous has , then for all , there must be an that maps to , so since the value above go from negative to positive twice, zero must be mapped to at least twice.

We notice that this function has 2 extrema which are located at x=-1 and x=0. We could have also of found this by looking at the function itself:

.

We know that extrema exist when the slope of the function is zero, hence we take the derivative, set it equal to zero, and than solve for x.

The derivative is the following:

.

Therefore setting both pieces equal to zero we see the following:

and .

Finding out if one of these x values produce a local min or max however requires either the first derivative test, the second derivative test, or analyzing the graph. We see that for a small neighborhood around , is the largest term, hence it is a local max. Similarly, for a small neighborhood around , is the smallest term, hence is a local min and the only local min of this function. Remember that it is not the x value that must be the smallest, rather it is it's corresponding y value.

The mean value theorem states that if continuous has , then for all , there must be an that maps to , so since the value above go from negative to positive twice, zero must be mapped to at least twice.

We notice that this function has 2 extrema which are located at x=-1 and x=0. We could have also of found this by looking at the function itself:

.

We know that extrema exist when the slope of the function is zero, hence we take the derivative, set it equal to zero, and than solve for x.

The derivative is the following:

.

Therefore setting both pieces equal to zero we see the following:

and .

Finding out if one of these x values produce a local min or max however requires either the first derivative test, the second derivative test, or analyzing the graph. We see that for a small neighborhood around , is the largest term, hence it is a local max. Similarly, for a small neighborhood around , is the smallest term, hence is a local min and the only local min of this function. Remember that it is not the x value that must be the smallest, rather it is it's corresponding y value.