# Video: AP Calculus AB Exam 1 β’ Section I β’ Part A β’ Question 18

The graph of π is shown in the figure. At which point is π continuous but not differentiable?

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### Video Transcript

The graph of π is shown in the figure. At which point is π continuous but not differentiable?

Letβs look in turn at each of the points π, π, π, and π on our graph of our function and establish what we do know about them. It seems sensible to begin with the point π. We can see we have this vertical dash line. This line will have the equation π₯ equals π, where π is some constant less than zero. If we look carefully at the graph of our function, we see that the line tends to π₯ equals π from the left and never quite reaches it. It also appears to do the same from the right. And this implies that the line π₯ equals π is an asymptote.

More importantly, we have a discontinuity here. It appears to be an asymptotic discontinuity though the type of discontinuity doesnβt actually matter. Weβre looking for a point on our curve which is continuous. So itβs absolutely canβt be continuous at this point. So weβll move on to point π. Itβs clearly continuous here. There are no jumps or holes in the curve. Thereβs also no reason to believe itβs not differentiable at this point either. We could add a tangent to the curve at this point. And this will allow us to easily find the rate of change of our function with respect to π₯ at this point. So it canβt be π. π is continuous and differentiable at this point.

Weβll now look at π. It does appear to be continuous though here we have this sharp turn in our graph. This means π canβt be differentiable at π. We canβt draw a unique tangent to the curve at this point. And this means π must therefore be continuous, but not differentiable at π. We will double check point π. We can see that we have a discontinuity. This open circle tells us that the function is not defined at this point. And so it canβt be continuous here.

π is continuous and not differentiable at the point π.