Video: AP Calculus AB Exam 1 β€’ Section I β€’ Part B β€’ Question 76

Let 𝑓 be the function given by 𝑓(π‘₯) = √(|π‘₯ βˆ’ 1|) for all π‘₯. Which of the following statements is true? [A] lim_(π‘₯ β†’ 1) 𝑓(π‘₯) β‰  0. [B] The graph has a vertical asymptote at π‘₯ = 1. [C] 𝑓 is both continuous and differentiable at π‘₯ = 1. [D] 𝑓 is continuous but not differentiable at π‘₯ = 1.

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Video Transcript

Let 𝑓 be the function given by 𝑓 of π‘₯ is equal to the square root of the modulus of π‘₯ minus one for all π‘₯. Which of the following statements is true? a) The limit as π‘₯ tends to one of 𝑓 of π‘₯ is not equal to zero. b) The graph has a vertical asymptote at π‘₯ equals one. c) 𝑓 is both continuous and differentiable at π‘₯ equals one. Or d) 𝑓 is continuous but not differentiable at π‘₯ equals one.

Let’s begin by establishing what we do know about our function 𝑓. It’s given by the square root of the modulus of π‘₯ minus one. In other words, we take the value of π‘₯ minus one, make sure it’s positive. And then, we find the square root of this value. So which of these statements is true?

For the first statement, let’s see what the limit of our function is as π‘₯ tends to one. Here, we don’t need to perform any manipulation of the algebraic expression. We can simply substitute π‘₯ equals one in. When we do, we see that the limit is equal to the square root of the modulus of one minus one. Well, one minus one is zero. And the modulus of zero is zero. So our limit is the square root of zero, which is itself zero. Statement a) said that the limit was not equal to zero. So this first statement can’t be true.

And we move on to part b). We’re told that the graph has a vertical asymptote at π‘₯ equals one. Remember, an asymptote is a line that a curve approaches but never touches. Let’s sketch our graph out. First, we recall that the graph of 𝑦 equals the square root of π‘₯ is the top or the positive half of the horizontal parabola π‘₯ equals 𝑦 squared. And that includes the point at the origin.

The graph of 𝑦 equals the square root of the modulus of π‘₯ includes this extra curve. That’s a reflection in the 𝑦-axis. We can then translate this curve one unit to the right. And we now have the graph for 𝑦 equals the square root of the modulus of π‘₯ minus one. Notice there simply cannot be a vertical asymptote at the point π‘₯ equals one. And as we’re not interested in testing for horizontal asymptotes, we’re done. Statement b) cannot be true.

But what about c) and d)? We’ll consider these together since they’re both asking whether the function is continuous and whether it’s differentiable at π‘₯ equals one. We can use our powers of deduction to assume that it must be continuous. But let’s check. There are no holes, jumps, or similar examples of discontinuities. But we do have this sharp turn at π‘₯ equals one. This means it cannot be differentiable at this point.

The correct answer here is d). Our function 𝑓 is continuous. But it’s not differentiable at π‘₯ equals one.

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