Vertical tangent lines If a function f is con...

Question

Vertical tangent lines If a function f is continuous at a and lim x→a| f′(x)|=∞, then the curve y=f(x) has a vertical tangent line at a, and the equation of the tangent line is x=a. If a is an endpoint of a domain, then the appropriate one-sided derivative (Exercises 71–72) is used. Use this information to answer the following questions.
73. {Use of Tech} Graph the following functions and determine the location of the vertical tangent lines.
c. f(x) = √|x-4|

73. {Use of Tech} Graph the following functions and determine the location of the vertical tangent lines.

c. f(x) = √|x-4|

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to sketch the graph of the function G of X, which is equal to 2 multiplied by the square root of the absolute value of X minus 7 and its vertical tangent line. So we're asked to create a sketch of our graph, which is for the function of GFX. So if we look at our function G of X, we noticed that it is actually defined for all values of X, since we have the absolute value sign underneath the square root sign. So that means we'll always have a positive quantity underneath the square root sign. We also see that G of X is always going to be positive, since we're going to be looking at the positive square root here. That means that G of X is going to be greater than or equal to 0. So, if we look at the point of when G of X is going to be at a minimum, when it is equal to 0, that occurs when the quantity inside the absolute value sign, the X minus 7, is equal to 0. So that means that we have a critical point at X equal to 7. And that G of 7 is going to be equal to 0. We're also gonna need several other points to plot in order to create our graph. So the first point we're going to look at is when X is equal to -2, we will calculate G of -2. And this is going to be equal to 2, multiplied by the square root of the absolute value of minus 2, minus 7. And when we evaluate this expression, this is going to be equal to 6. Next point we'll look at is when X is equal to 3, so we'll look for GF 3. And this is going to be equal to 2, multiplied by the square root of the absolute value of 3 minus 7, and this evaluates to 4. The next point we'll look at is, um, when X is equal to 6, so we'll calculate G of 6. So this is going to be equal to 2, multiplied by the square root of the absolute value of 6 minus 7, and when we evaluate this expression, this is going to be equal to 2. We'll also need to uh to calculate a few more points. So we're going to look at G of 8, so when X is equal to 8. And this is going to be equal to 2 multiplied by the square root of the absolute value of 8 minus 7. And this is going to be equal to 2 when we evaluate that expression. The next point we'll look at is when X is equal to 11, we'll calculate G of 11. This is going to be equal to 2, multiplied by the square root of the absolute value of 11 minus 7, and this is going to be equal to 4. And finally, we'll look at the points when um X is equal to 16, so we'll calculate G of 16. And this is going to be equal to 2 multiplied by the square root of the absolute value of 16 minus 7, and this is going to be equal to 6 when we evaluate that expression. So now we have several points to actually plot on our graph, but we also need to determine where the vertical tangent line is located. And so to do this, we need to take a derivative of our function GFX. To make it a little bit easier to take that derivative, we're going to rewrite our function G of X as a piece-wise function. We'll have GFX. It's going to be equal to. 2 multiplied by the square root of the quantity of 7 minus X. When X is less than or equal to 7, And to multiply by the square root of the quantity of X minus 7, when X is greater than 7. So this will ensure that the quantity underneath the square root sign is always positive. So when we take a derivative of this function, we'll be looking for G of X. And this is going to be equal to. And when we take the derivative of the first piece of our piece-wise function, we will have -1, divided by the square root of the quantity of 7 minus X, and that is 4 X less than or equal to 7. And when we take the derivative of the second piece of our piece-wise function of G of X, that is going to give us 1 divided by the square root of the quantity of X minus 7, and that is for X greater than 7. So if we look at both pieces of our piece-wise function 4G X, we notice that the denominator is equal to 0 in both cases when X is equal to 7. And since I have a 0 on the denominator here, that means our function is not defined. So our derivative is not defined at X equal to 7, and that is one of the indications that we have a vertical tangent line at X equals to 7. So our vertical tangent line is located at X equal to 7. So now we have enough information to create our graph. So the first thing we're going to do is to plot our points on our graph. So, we're gonna stop by um plotting the point of -2.6. Next point is going to be 3.4. The next point is going to be 6.2. Next point is going to be 7.0. Next point is 8.2. Next point is 11.4. And the last point is 16.6. So now what we're going to do is to connect these points with a smooth curve. Being sure to go through every single point as we pass through. The various points, and we noticed that we come down to a sharp point when X is equal to 7. So that's also an indication that we have a vertical tangent line at that point. And so making sure that we go through all of the points. On our graph. And now we need to actually draw our vertical tangent lines. We'll draw that as a dash line at X equal to 7. So this is the graph that the problem was asking us to create. So this is the answer to the problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Vertical Tangent Lines

Continuity of Functions

One-Sided Derivatives

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