Video: Using the Graph of a Function to Determine the Sign of Its First and Second Derivatives

At which point on the given graph of 𝑦 = 𝑓(π‘₯) is 𝑓′(π‘₯) > 0 and 𝑓″(π‘₯) < 0?

04:16

Video Transcript

At which point on the given graph of 𝑦 equals 𝑓 of π‘₯ is 𝑓 prime of π‘₯ greater than zero and 𝑓 double prime of π‘₯ less than zero?

So, we’ve been given the graph of a function and asked to use this to determine something about the sign of both its first and second derivatives at various points. Let’s use a simple table to organize our thoughts. Let’s begin with the first condition, which is that 𝑓 prime of π‘₯ must be greater than zero β€” that’s positive β€” at the point we’re looking for.

We recall, first of all, then that a function’s first derivative gives the slope of the curve or in fact the slope of the tangent to the curve at any point. So, by sketching in tangents to the graph of 𝑦 equals 𝑓 of π‘₯ at the four points, we will be able to determine whether they slope upwards, meaning 𝑓 prime of π‘₯ is positive, or downwards, meaning 𝑓 prime of π‘₯ is negative.

For example, at point A, we can see that the tangent slopes upwards from left to right. And therefore, 𝑓 prime of π‘₯ is positive at point A. However, at point B, the tangent slopes downwards from left to right. And therefore, 𝑓 prime of π‘₯, the first derivative, is negative at point B. At point C, the tangent once again slopes upwards. So, the first derivative is positive here. And at point D, it looks as if the tangent to the curve is completely flat. It’s horizontal. In which case, 𝑓 prime of π‘₯, the first derivative, will be equal to zero at point D. So, we find that, of the four points, there are only two at which the first derivative is positive: point A and point C.

Now, let’s consider the second condition, which is that the second derivative, 𝑓 double prime of π‘₯, must be negative at the point we’re looking for. We recall that the second derivative is the derivative of the first derivative. And so, if the second derivative is negative, then this means that the first derivative will be decreasing. This is related to the shape or concavity of the graph of 𝑓 of π‘₯ itself. In the sketch here, we see that the slope of the tangent is changing from positive to zero to negative. And so, its value is decreasing. And a curve shaped like this is referred to as being concave down. The reverse of this is also true. When a function’s second derivative is positive, then its first derivative is increasing. And the function is referred to as being concave up.

What we also notice that in a region where a graph is concave down, then the tangents to the curve lie above the curve itself. So, we can use the relative positioning of the tangents to determine whether a graph is concave up or down and, hence, whether its second derivative is positive or negative. Looking back at our original graph then, we see that at point A, the tangent to the curve lies above the curve itself. And so, the graph is concave down at point A. And the second derivative, 𝑓 double prime of π‘₯, will be negative at this point. At point B, the same is true. The tangent lies above the curve. The graph is still concave down. And therefore, the second derivative is still negative.

However, at point C, the tangent to the curve lies below the curve itself, meaning the graph is concave up at point C. And so, the second derivative, 𝑓 double prime of π‘₯, will be positive here. Finally, at point D, the tangent once again lies above the curve, meaning the curve is concave down and the second derivative will once again be negative at point D.

Of the four points then, there are three points at which the second derivative is negative: point A, point B, and point D. The only point which satisfies both criteria is point A. So, by considering the slope of the tangents to reveal the sign of the first derivative and then the concavity of the curve to reveal the sign of the second derivative, we’ve found that the only point at which 𝑓 prime of π‘₯ is positive, but 𝑓 double prime of π‘₯ is negative is point A.

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