Corresponding characteristics of the graphs of ƒ, ƒ', and ƒ'' - AP Calculus AB

Card 0 of 33

Question

$h(x)=\frac{f+g}{f}$

If

$f^{'}(x)=2$

and $g^{'}(x)=4$ ,

then find $h^{'}(x)$ .

Answer

We see the answer is when we use the product rule.

$h(x)=\frac{f+g}{f}$

$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}}$

$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}} =-2$

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Question

On what intervals is the function both concave up and decreasing?

Answer

The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get

Solving for the zero's, we see hits zero at and . Constructing an interval test,

$(-\infty, -3), (-3, 1/3), (1/3, \infty)$ , we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.

Thus, we can see that the derivative is only negative on the interval .

Repeating the process for the second derivative,

The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are

$(-\infty, -4/3), (-4/3, \infty)$ . Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.

Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:

$(-3, 1/3) \cap (-4/3, \infty) = (-4/3, 1/3)$

If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.

Thus, our final answer is

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Question

On a closed interval, the function is decreasing. What can we say about and on these intervals?

Answer

If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.

For to be decreasing, would have to be negative, which we don't know.

being negative has nothing to do with its slope.

For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.

Thus, the only correct answer is that is negative.

Compare your answer with the correct one above

Question

$h(x)=\frac{f+g}{f}$

If

$f^{'}(x)=2$

and $g^{'}(x)=4$ ,

then find $h^{'}(x)$ .

Answer

We see the answer is when we use the product rule.

$h(x)=\frac{f+g}{f}$

$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}}$

$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}} =-2$

Compare your answer with the correct one above

Question

On what intervals is the function both concave up and decreasing?

Answer

The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get

Solving for the zero's, we see hits zero at and . Constructing an interval test,

$(-\infty, -3), (-3, 1/3), (1/3, \infty)$ , we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.

Thus, we can see that the derivative is only negative on the interval .

Repeating the process for the second derivative,

The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are

$(-\infty, -4/3), (-4/3, \infty)$ . Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.

Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:

$(-3, 1/3) \cap (-4/3, \infty) = (-4/3, 1/3)$

If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.

Thus, our final answer is

Compare your answer with the correct one above

Question

On a closed interval, the function is decreasing. What can we say about and on these intervals?

Answer

If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.

For to be decreasing, would have to be negative, which we don't know.

being negative has nothing to do with its slope.

For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.

Thus, the only correct answer is that is negative.

Compare your answer with the correct one above

Question

$h(x)=\frac{f+g}{f}$

If

$f^{'}(x)=2$

and $g^{'}(x)=4$ ,

then find $h^{'}(x)$ .

Answer

We see the answer is when we use the product rule.

$h(x)=\frac{f+g}{f}$

$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}}$

$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}} =-2$

Compare your answer with the correct one above

Question

On what intervals is the function both concave up and decreasing?

Answer

The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get

Solving for the zero's, we see hits zero at and . Constructing an interval test,

$(-\infty, -3), (-3, 1/3), (1/3, \infty)$ , we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.

Thus, we can see that the derivative is only negative on the interval .

Repeating the process for the second derivative,

The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are

$(-\infty, -4/3), (-4/3, \infty)$ . Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.

Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:

$(-3, 1/3) \cap (-4/3, \infty) = (-4/3, 1/3)$

If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.

Thus, our final answer is

Compare your answer with the correct one above

Question

On a closed interval, the function is decreasing. What can we say about and on these intervals?

Answer

If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.

For to be decreasing, would have to be negative, which we don't know.

being negative has nothing to do with its slope.

For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.

Thus, the only correct answer is that is negative.

Compare your answer with the correct one above

Question

$h(x)=\frac{f+g}{f}$

If

$f^{'}(x)=2$

and $g^{'}(x)=4$ ,

then find $h^{'}(x)$ .

Answer

We see the answer is when we use the product rule.

$h(x)=\frac{f+g}{f}$

$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}}$

$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}} =-2$

Compare your answer with the correct one above

Question

On what intervals is the function both concave up and decreasing?

Answer

The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get

Solving for the zero's, we see hits zero at and . Constructing an interval test,

$(-\infty, -3), (-3, 1/3), (1/3, \infty)$ , we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.

Thus, we can see that the derivative is only negative on the interval .

Repeating the process for the second derivative,

The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are

$(-\infty, -4/3), (-4/3, \infty)$ . Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.

Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:

$(-3, 1/3) \cap (-4/3, \infty) = (-4/3, 1/3)$

If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.

Thus, our final answer is

Compare your answer with the correct one above

Question

On a closed interval, the function is decreasing. What can we say about and on these intervals?

Answer

If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.

For to be decreasing, would have to be negative, which we don't know.

being negative has nothing to do with its slope.

For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.

Thus, the only correct answer is that is negative.

Compare your answer with the correct one above

Question

$h(x)=\frac{f+g}{f}$

If

$f^{'}(x)=2$

and $g^{'}(x)=4$ ,

then find $h^{'}(x)$ .

Answer

We see the answer is when we use the product rule.

$h(x)=\frac{f+g}{f}$

$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}}$

$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}} =-2$

Compare your answer with the correct one above

Question

On what intervals is the function both concave up and decreasing?

Answer

The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get

Solving for the zero's, we see hits zero at and . Constructing an interval test,

$(-\infty, -3), (-3, 1/3), (1/3, \infty)$ , we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.

Thus, we can see that the derivative is only negative on the interval .

Repeating the process for the second derivative,

The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are

$(-\infty, -4/3), (-4/3, \infty)$ . Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.

Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:

$(-3, 1/3) \cap (-4/3, \infty) = (-4/3, 1/3)$

If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.

Thus, our final answer is

Compare your answer with the correct one above

Question

On a closed interval, the function is decreasing. What can we say about and on these intervals?

Answer

If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.

For to be decreasing, would have to be negative, which we don't know.

being negative has nothing to do with its slope.

For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.

Thus, the only correct answer is that is negative.

Compare your answer with the correct one above

Question

$h(x)=\frac{f+g}{f}$

If

$f^{'}(x)=2$

and $g^{'}(x)=4$ ,

then find $h^{'}(x)$ .

Answer

We see the answer is when we use the product rule.

$h(x)=\frac{f+g}{f}$

$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}}$

$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}} =-2$

Compare your answer with the correct one above

Question

On what intervals is the function both concave up and decreasing?

Answer

The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get

Solving for the zero's, we see hits zero at and . Constructing an interval test,

$(-\infty, -3), (-3, 1/3), (1/3, \infty)$ , we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.

Thus, we can see that the derivative is only negative on the interval .

Repeating the process for the second derivative,

The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are

$(-\infty, -4/3), (-4/3, \infty)$ . Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.

Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:

$(-3, 1/3) \cap (-4/3, \infty) = (-4/3, 1/3)$

If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.

Thus, our final answer is

Compare your answer with the correct one above

Question

On a closed interval, the function is decreasing. What can we say about and on these intervals?

Answer

If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.

For to be decreasing, would have to be negative, which we don't know.

being negative has nothing to do with its slope.

For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.

Thus, the only correct answer is that is negative.

Compare your answer with the correct one above

Question

$h(x)=\frac{f+g}{f}$

If

$f^{'}(x)=2$

and $g^{'}(x)=4$ ,

then find $h^{'}(x)$ .

Answer

We see the answer is when we use the product rule.

$h(x)=\frac{f+g}{f}$

$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}}$

$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}} =-2$

Compare your answer with the correct one above

Question

On what intervals is the function both concave up and decreasing?

Answer

The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get

Solving for the zero's, we see hits zero at and . Constructing an interval test,

$(-\infty, -3), (-3, 1/3), (1/3, \infty)$ , we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.

Thus, we can see that the derivative is only negative on the interval .

Repeating the process for the second derivative,

The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are

$(-\infty, -4/3), (-4/3, \infty)$ . Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.

Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:

$(-3, 1/3) \cap (-4/3, \infty) = (-4/3, 1/3)$

If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.

Thus, our final answer is

Compare your answer with the correct one above

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