High School Math : Calculus I — Derivatives

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #2 : Integrals

The speed of a car traveling on the highway is given by the following function of time:

Note that

What does this mean?

Possible Answers:

The car's speed is constantly changing at time .

The car is not accelerating at time .

The car takes  seconds to reach its maximum speed.

The car is not decelerating at time .

The car is not moving at time .

Correct answer:

The car is not moving at time .

Explanation:

The function  gives you the car's speed at time . Therefore, the fact that  means that the car's speed is  at time . This is equivalent to saying that the car is not moving at time . We have to take the derivative of  to make claims about the acceleration.

Example Question #11 : Calculus I — Derivatives

The speed of a car traveling on the highway is given by the following function of time:

Consider a second function:

What can we conclude about this second function?

Possible Answers:

It represents the rate at which the speed of the car is changing.

It represents the change in distance over a given time .

It has no relation to the previous function.

It represents the total distance the car has traveled at time .

 

It represents another way to write the car's speed.

Correct answer:

It represents the rate at which the speed of the car is changing.

Explanation:

Notice that the function  is simply the derivative of  with respect to time. To see this, simply use the power rule on each of the two terms. 

Therefore,  is the rate at which the car's speed changes, a quantity called acceleration.

Example Question #1 : Applications Of Derivatives

Define .

Give the interval(s) on which  is decreasing.

Possible Answers:

Correct answer:

Explanation:

 is decreasing on those intervals at which .

 

 

We need to find the values of  for which . To that end, we first solve the equation:

 

These are the boundary points, so the intervals we need to check are:

,  and 

 

We check each interval by substituting an arbitrary value from each for .

 

Choose 

 increases on this interval.

 

Choose 

 decreases on this interval.

 

Choose 

 increases on this interval.

 

The answer is that  decreases on .

Example Question #2 : Applications Of Derivatives

Define .

Give the interval(s) on which  is increasing.

Possible Answers:

Correct answer:

Explanation:

 is increasing on those intervals at which .

 

 

We need to find the values of  for which . To that end, we first solve the equation:

 

These are the boundary points, so the intervals we need to check are:

,  and 

We check each interval by substituting an arbitrary value from each for .

 

Choose 

 increases on this interval.

 

Choose 

 decreases on this interval.

 

Choose 

 increases on this interval.

 

The answer is that  increases on 

Example Question #1 : Applications Of Derivatives

At what point does  shift from increasing to decreasing?

Possible Answers:

It does not shift from increasing to decreasing

Correct answer:

Explanation:

To find out where the graph shifts from increasing to decreasing, we need to look at the first derivative. 

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat  as  since anything to the zero power is one.

Notice that  since anything times zero is zero.

If we were to graph , would the y-value change from positive to negative? Yes. Plug in zero for y and solve for x.

Example Question #21 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

At what point does  shift from decreasing to increasing?

Possible Answers:

Correct answer:

Explanation:

To find out where it shifts from decreasing to increasing, we need to look at the first derivative. The shift will happen where the first derivative goes from a negative value to a positive value.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

Remember that anything to the zero power is one.

Can this equation be negative? Yes. Does it shift from negative to positive? Yes. Therefore, it will shift from negative to positive at the point that .

Example Question #1241 : Ap Calculus Ab

At what value of  does  shift from decreasing to increasing?

Possible Answers:

It does not shift from decreasing to increasing

Correct answer:

Explanation:

To find out when the function shifts from decreasing to increasing, we look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

Anything to the zero power is one.

From here, we want to know if there is a point at which graph changes from negative to positive. Plug in zero for y and solve for x.

This is the point where the graph shifts from decreasing to increasing.

 

Example Question #11 : Calculus I — Derivatives

At the point , is the function  increasing or decreasing, concave or convex?

Possible Answers:

Decreasing, convex

The function is undefined at that point

Increasing, convex

Decreasing, concave

Increasing, concave

Correct answer:

Decreasing, convex

Explanation:

First, let's find out if the graph is increasing or decreasing. For that, we need the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat  as  since anything to the zero power is one.

Notice that  since anything times zero is zero.

Plug in our given point for . If the result is positive, the function is increasing. If the result is negative, the function is decreasing.

Our result is negative, therefore the function is decreasing.

To find the concavity, look at the second derivative. If the function is positive at our given point, it is concave. If the function is negative, it is convex.

To find the second derivative we repeat the process, but using  as our expression.

As you can see, our second derivative is a constant. It doesn't matter what point we plug in for ; our output will always be negative. Therefore our graph will always be convex.

Combine our two pieces of information to see that at the given point, the graph is decreasing and convex.

Example Question #2 : Finding Regions Of Concavity And Convexity

Let . What is the largest interval of x for which f(x) is concave upward?

Possible Answers:

Correct answer:

Explanation:

This question asks us to examine the concavity of the function . We will need to find the second derivative in order to determine where the function is concave upward and downward. Whenever its second derivative is positive, a function is concave upward.

Let us begin by finding the first derivative of f(x). We will need to use the Product Rule. According to the Product Rule, if , then . In this particular problem, let  and . Applying the Product rule, we get

In order to evaluate the derivative of , we will need to invoke the Chain Rule. According to the Chain Rule, the derivative of a function in the form  is given by . In finding the derivative of , we will let  and .

We can now finish finding the derivative of the original function. 

 

To summarize, the first derivative of the funciton  is .

 We need the second derivative in order to examine the concavity of f(x), so we will differentiate one more time. Once again, we will have to use the Product Rule in conjunction with the Chain Rule.

In order to find where f(x) is concave upward, we must find where f''(x) > 0.

In order to solve this inequality, we can divide both sides by . Notice that  is always positive (because e raised to any power will be positive); this means that when we divide both sides of the inequality by , we won't have to flip the sign. (If we divide an inequality by a negative quantity, the sign flips.)

Dividing both sides of the inequality by  gives us

When solving inequalities with polynomials, we often need to factor.

Notice now that the expression  will always be positive, because the smallest value it can take on is 3, when x is equal to zero. Thus, we can safely divide both sides of the inequality by  without having to change the direction of the sign. This leaves us with the inequality

, which clearly only holds when .

Thus, the second derivative of f''(x) will be positive (and f(x) will be concave up) only when . To represent this using interval notation (as the answer choices specify) we would write this as .

The answer is .

 

 

Example Question #12 : Calculus I — Derivatives

At the point , is  increasing or decreasing, and is it concave or convex?

Possible Answers:

Increasing, convex

Decreasing, concave

Decreasing, convex

The graph is undefined at point 

Increasing, concave

Correct answer:

Decreasing, convex

Explanation:

To find out if the function is increasing or decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

Anything to the zero power is one.

Now we plug in our given value and find out if the result is positive or negative. If it is positive, the function is increasing. If it is negative, the function is decreasing.

Therefore, the function is decreasing.

To find out if it is concave or convex, look at the second derivative. If the result is positive, it is convex. If it is negative, then it is concave.

To find the second derivative, we repeat the process using  as our expression.

We're going to treat  as .

Notice that  since anything times zero is zero.

As stated before, anything to the zero power is one.

Since we get a positive constant, it doesn't matter where we look on the graph, as our second derivative will always be positive. That means that this graph is going to be convex at our given point.

Therefore, the function is decreasing and convex at our given point.

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