Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 61 and 62, the graph of f' is given. Assume that f is continuous, and determine the x-values corresponding to local minima and local maxima.

Graph of a blue curve labeled f prime showing changes in slope with x and y axes marked from -4 to 4.

Accepted Answer

Welcome back, everyone. Find the X coordinates of the local minima and maxima of the continuous function F of X given the graph of its derivative F of X. Before we analyze the given graph, let's recall how we can use derivatives to identify the local minima and maxima. So first of all, we have to recall that we're looking for the change in sign of the derivative. For example, if we had a positive derivative and it became negative, it basically indicates that the function was increasing. And then it started to decrease, right? So it shows us that we have a local maximum. Alternatively, if we have a negative derivative which becomes positive, it indicates that the function was decreasing and started to increase. This shows us that we have a local minimum. So considering the graph, we're looking for the intercept points, right, because we're given the first derivative. The first X intercept that we can see is X equals -2. We also have an intercept at X equals 2.5, and the final third intercept is at X equals 5. What can we tell about the sign of the derivative for X? That is less than -2. We can see that the derivative is positive. Then for X. That is greater than -2. But less than 2.5. We can conclude that the derivative is negative because now it is below the X-axis. So initially, the derivative was positive, the function was increasing, and then it starts to decrease, right? So it was increasing up to -2, and then started to decrease up to 2.5, right? But after 2.5, we still have a negative value of the derivative until we reach 5, so we can actually conclude that this is true for X being less than 5. So now looking at our diagram, we can see that the function was increasing before -2, and then started to decrease after -2, meaning we have a local maximum. So we're gonna conclude that we have a local maximum. When X is equal to -2, because we have an intercept, where the derivative changes a sign from being positive to being negative. Then we already know that X equals 2.5 is not a local. Extreme. Point And that's because the derivative doesn't change its sign. Yes, we do have an intercept, but the derivative was negative before 2.5, and after 2.5, it is still negative, right? So the derivative does not change its sign, the function is still decreasing. Now, what about 5? Well, we can see that for X, That is less than 5. But greater than -2. The derivative is negative. And then whens. Greater than 5, the derivative becomes positive. We have an intercept. Where the derivative changes a sign from negative to positive, so it indicates that the function was decreasing. And started to increase after x equals 5. So this indicates a local minimum. We can conclude that we have a local minimum. A X equals 5. And those will be our answers. We have a local maximum at X equals -2, and a local minimum at X equals 5. Thank you for watching.

Identifying Extrema

In Exercises 61 and 62, the graph of f' is given. Assume that f is continuous, and determine the x-values corresponding to local minima and local maxima.

Key Concepts

Critical Points and Extrema

First Derivative Test

Interpreting the Graph of f'

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