Calculus I — Derivatives - Math

For this problem, we can use the power rule. The power rule states that we multiply each variable by its current exponent and then lower that exponent by one.

Simplify.

Anything to the zero power is one, so .

Therefore, .

To find potential minima of the function, take the first derivative of using the power rule.

Set the derivative to 0:

We solve for to obtain and then plug in 0.5 into the original function to obtain the answer of

We can double check that is indeed a minimum by using the second derivative test

which means the function is concave up, so that the point we found is a minimum.

To find the maximum, we need to look at the first derivative.

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

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

Notice that since anything times zero is zero.

When looking at the first derivative, remember that if the output of this equation is positive, the original function is increasing. If the derivative is negative, then the function is decreasing.

Notice that changes from positive to negative when .

We can find that root using the quadratic equation:

Since we're looking for a negative value, we will subtract.

Therefore, the maximum is at .

To find the maximum, we need to look at the first derivative.

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

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

Notice that since anything times zero is zero.

When looking at the first derivative, remember that if the output of this equation is positive, the original function is increasing. If the derivative is negative, then the function is decreasing.

Because we want the MINIMUM, we want to see where the derivative changes from negative to positive.

Notice that has a root when . In fact, it changes from negative to positive at that particular point. This is the local minimum in the interval .

To find the maximum, we must find where the graph shifts from increasing to decreasing. To find out the rate at which the graph shifts from increasing to decreasing, we look at the second derivative and see when the value changes from positive to negative.

That is to say, we will look at the second derivative and see where (if at all) the graph crosses the x-axis and is moving from a positive y-value to a negative y-value.

Now we must find the second derivative. Unfortunately, the derivatives of trig functions must be memorized. The first derivative is:

.

To find the second derivative, we take the derivative of our result.

.

Therefore, the second derivative will be .

Does our new equation cross the x-axis and move from positive to negative between and ? Yes. It does once, when . Therefore, our local maximum shall be when . Plug that value back into our first equation to find that the maximum will be at the point .

A local minimum occurs when a graph "bottoms out" -- it has been decreasing, it slows down, stops, and then begins to increase. At that point when it switches from decreasing to increasing, our first derivative should move from negative to positive. Start by finding the first derivative and then see if that happens.

To find the first derivative of , we can use the power rule.

The power rule states that we multiply each variable by its current exponent and then lower the exponent of each variable by one.

Since , we're going to treat as .

Anything times zero is zero, so our final term , regardless of the power of the exponent.

Simplify what we have.

Our first derivative, then, is .

Graph the equation . Does it move from negative to positive when ? Yes it does. Therefore, that zero point will be our minimum.

The zero occurs when . Therefore the minimum of our original graph is .

To find the local minimum of , we need to look at the first derivative.

To find the first derivative, we can use the power rule. The power rule states that we multiply each variable by its current exponent and then lower that exponent by one.

Simplify.

Anything to the zero power is one, so .

Therefore, .

At the minimum, our graph will cross the -axis. Therefore, we need to find the roots. Use the quadratic equation:

From here we split off into two roots, one where we add and one where we subtract:

and

Do both of these roots satisfy ? Yes.

Then we move on to our next question: Does the graph shift from negative to positive at either of these roots? Yes. When , the graph shifts from negative to positive.

Therefore, the local minimum is at .

To find the minimum we need to look at the first derivative.

Since we're adding terms, we take the derivative of each part separately. For , we can use the power rule, which states that we multiply the variable by the current exponent and then lower the exponent by one. For sine, we use our trigonometric derivative rules.

Remember, .

Now we need to find the roots of the derivative.

Does cross the -axis? Yes, it crosses at .

Our next question is, "Does the graph change from negative to positive at that point?" Yes. This means that our parent function has shifted from decreasing to increasing at that point.

Therefore, this will be our minimum.

To find the slope of a line tangent to a parabola at a specific point, find the derivative of the parabola's equation, then substitute the -coordinate of the specific point in the new equation.

In this case, it helps to expand the equation before taking the derivative:

Now take the derivative of the expanded equation:

Since the -intercept is the point where the -coordinate is , substitute into the equation for .

You can substitute to write this as:

Note that as ,

, since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

, which is the correct choice.

Substitute to rewrite this limit in terms of u instead of x. Multiply the top and bottom of the fraction by 2 in order to make this substitution:

(Note that as , .)

, so

, which is therefore the correct answer choice.

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.

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.

To find the concavity, we need to look at the first and second derivatives at the given point.

To take the first derivative of this equation, use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent:

Simplify:

Remember that anything to the zero power is equal to one.

The first derivative tells us if the function is increasing or decreasing. Plug in the given point, , to see if the result is positive (i.e. increasing) or negative (i.e. decreasing).

Therefore the function is increasing.

To find out if the function is convex, we need to look at the second derivative evaluated at the same point, , and check if it is positive or negative.

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

Notice that since anything times zero is zero.

Plug in our given value:

Since the second derivative is positive, the function is convex.

Therefore, we are looking at a graph that is both increasing and convex at our given point.

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.

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.

To find if the equation is increasing or decreasing, we need to look at the first derivative. If our result is positive at , then the function is increasing. If it is negative, then the function is decreasing.

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.

Plug in our given value.

Is it positive? Yes. Then it is increasing.

To find the concavity, we need to look at the second derivative. If it is positive, then the function is concave up. If it is negative, then the function is concave down.

Repeat the process we used for the first derivative, but use as our expression.

For this problem, we're going to say that since, as stated before, anything to the zero power is one.

Notice that as anything times zero is zero.

As you can see, there is no place for a variable here. It doesn't matter what point we look at, the answer will always be positive. Therefore this graph is always concave up.

This means that at our given point, the graph is increasing and concave up.

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 .

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