Does the graph of U(x)={ 0, x<0 1, x ≥ 0 . have a vertical tangent at the point (0,1

Does the graph of U(x)={ 0, x<0 1, x ≥ 0 . have...

Key Concepts

Vertical Tangents

A vertical tangent occurs on a graph where the slope of the tangent line becomes infinitely steep, usually when the derivative tends toward infinity or negative infinity. This phenomenon implies that the function’s rate of change is unbounded at that point, resulting in a vertical line as the tangent.

Continuity

Continuity is a necessary condition for a function to have a well?defined tangent line at a point. If a function is discontinuous at a point, the concept of a tangent line (vertical or otherwise) does not apply because the function does not smoothly connect at that location.

Jump Discontinuity

A jump discontinuity is a type of discontinuity where the function makes an abrupt change in value at a particular point. Since the graph 'jumps' from one value to another, there is no single point of contact through which a tangent can be drawn, distinct from the concept of a vertical tangent which requires the function to be continuous.

Transcript

00:01 In this problem, we're asked of the function u of x equal to zero when x is less than zero, and one, when x is bigger than equal to zero, has a vertical tangent line at the point zero, comma, one.

00:12 Before we actually do this calculation, let's just sketch what this graph looks like.

00:16 We're not actually going to need to know what the graph is to do this problem, but it can be kind of helpful as a visualization.

00:22 When x is less than zero, we have a flat line at zero, but then when x is bigger than zero, the line jumps to one.

00:30 So this is the graph that we're working with.

00:33 And so what we need to do to check if this has a vertical tangent line at zero is we should try and calculate the limit as h goes to 0 of u of x plus h minus u of x all over h.

00:48 And we should see if this limit is actually exists or not.

00:52 Remember we're doing this at the point x equals 0.

00:55 So we can make this substitution, the limit of h goes to 0 of u of h minus u of 0.

01:00 Which is 1, all over h.

01:05 So now we need to evaluate this limit and ask ourselves, does this exist? so we should see like do we get an infinity or a minus infinity here? and if so, then we do have a vertical tangent line.

01:17 Otherwise we don't.

01:19 Now this is a bit tricky to evaluate because our u function is different depending on if h is positive or if h is negative.

01:27 Because our u of x is piecewise, we're going to have to separate our limit into the left -hand limit and the right -hand limit...

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