Introductory Calculus - Pre-Calculus

To solve, simply realize a constant derived is always 0. Thus, the answer is 0.

We begin by recalling that one way of defining the derivative of a function is the slope of the tangent line of the function at a given point. Therefore, finding the derivative of our equation will allow us to find the slope of the tangent line. Since the two things needed to find the equation of a line are the slope and a point, we would be halfway done.

We calculate the derivative using the power rule.

However, we don't want the slope of the tangent line at just any point but rather specifically at the point . To obtain this, we simply substitute our x-value 1 into the derivative.

Therefore, the slope of our tangent line is .

We now need a point on our tangent line. Our choices are quite limited, as the only point on the tangent line that we know is the point where it intersects our original graph, namely the point .

Therefore, we can plug these coordinates along with our slope into the general point-slope form to find the equation.

Solving for will give us our slope-intercept form.

The equation of the tangent line at depends on the derivative at that point and the function value.

The derivative at that point of is

using the Power Rule

which means

The derivative is zero, so the tangent line will be horizontal.

It intersects it at since , so that line is .

Rewrite in slope-intercept form, , to determine the slope.

The slope of the given function is 2.

Substitute the slope and the given point, , in the slope-intercept form to determine the y-intercept.

Substitute this and the slope back to the slope-intercept equation.

The equation of the tangent line is:

We begin by finding the equation of the derivative using the limit definition:

We define and as follows:

We can then define their difference:

Then, we divide by h to prepare to take the limit:

Then, the limit will give us the equation of the derivative.

Now, we must realize that the slope of the line tangent to the curve at the given point is equivalent to the derivative at the point. So if we define our tangent line as: , then this m is defined thus:

Therefore, the equation of the line tangent to the curve at the given point is:

First, find the slope of this tangent line by taking the derivative:

Plugging in 1 for x:

So the slope is 4

Now we need to find the y-coordinate when x is 1, so plug 1 in to the original equation:

To write the equation, use point-slope form and then use algebra to change to slope-intercept like the answer choices:

distribute the 4

add 2 to both sides

First, find the slope of the tangent line by taking the first derivative:

To finish determining the slope, plug in the x-value, 2:

the slope is 6

Now find the y-coordinate where x is 2 by plugging in 2 to the original equation:

To write the equation, start in point-slope form and then use algebra to get it into slope-intercept like the answer choices.

distribute the 6

add 8 to both sides

First, take the first derivative in order to find the slope:

To continue finding the slope, plug in the x-value, -2:

Then find the y-coordinate by plugging -2 into the original equation:

The y-coordinate is

Now write the equation in point-slope form then algebraically manipulate it to match one of the slope-intercept forms of the answer choices.

distribute the -5

add to both sides

First distribute the . That will make it easier to take the derivative:

Now take the derivative of the equation:

To find the slope, plug in the x-value -3:

To find the y-coordinate of the point, plug in the x-value into the original equation:

Now write the equation in point-slope, then use algebra to get it into slope-intercept like the answer choices:

distribute

subtract from both sides

write as a mixed number

First find the slope of the tangent to the line by taking the derivative.

Using the Exponential Rule we get the following,

.

Then plug 1 into the equation as 1 is the point to find the slope at.

.

One way of finding the slope at a given point is by finding the derivative. In this case, we can take the derivative of y with respect to x, and plug in the desired value for x.

Using the exponential rule we get the following derivative,

.

Plugging in x=2 from the point 2,3 gives us the final slope,

Thus our slope at the specific point is .

Note that in this case, using the y coordinate was not necessary.

To find the slope of the tangent line of the function at the given value, evaluate the first derivative for the given.

The first derivative is

and for this function

and

So the slope is

To find the slope of the tangent line of the function at the given value, evaluate the first derivative for the given value.

The first derivative is

and for this function

and plugging in the specific x value we get,

So the slope is

.

Calculate the derivative of by using the derivative rules. The derivative function determines the slope at any point of the original function.

The derivative is:

With the given point , . Substitute this value to the derivative function to determine the slope at that point.

The slope of the tangent line that intersects point is .

Since the factor is in the numerator and the denominator, there is a removable discontinuity at . The function is not defined at , but function would move towards the same point for the resultant function .

Since the factor cannot be factored out, there is an infinite discontinuity at . The denominator will get very small and the numerator will move toward a fixed value.

There is no discontinuity at all at . The function simply evaluates to zero at this point.

The function in the numerator factors to:

so if we cancel the x+3 in the numerator and denominator we have the same function but it is continuous. The gives us a hole at x=-3 so our function is not continuous at x=-3.

In order to determine if a function is continuous at a point three things must happen.

1. Taking the limit from the lefthand side of the function towards a specific point exists.

2. Taking the limit from the righthand side of the function towards a specific point exists.

3. The limits from 1) and 2) are equal and equal the value of the original function at the specific point in question.

In our case,

Because all of these conditions are met, the function is continuous at 0.

First, find that at any point where , .

Then find that

and .

As these are all equal, it can be determined that the function is continuous on all points of its domain.

As approaches , the function approaches , which is undefined. However, if we factor , we get:

The factors in the numerator and denominator cancel out, leaving .

Therefore, our function is continuous at all values of from .

This is a graph of . We know that is undefined; therefore, there is no value for . But as we take a look at the graph, we can see that as approaches 0 from the left, approaches negative infinity.

This can be illustrated by thinking of small negative numbers.

NOTE: Pay attention to one-sided limit specifications, as it is easy to pick the wrong answer choice if you're not careful.

is actually infinity, not negative infinity.