Theory and Examples

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Question

Theory and Examples

[Technology Exercise] Graph the functions in Exercises 63–66. Then find the extreme values of the function on the interval and say where they occur.

f(x) = |x − 2| + |x + 3|, −5 ≤ x ≤ 5

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to graph the function and determine its absolute extreme values, and we're given the function F of X is equal to the absolute value of the quantity of X minus 1 in quantity, plus the absolute value of the quantity of X plus 5 in quantity. And we're given the interval of minus 7 is less than or equal to x, which is less than or equal to 5. And below the problem, we're given an empty graph on which to plot our function. So, the first thing we want to do is to plot our function F of X, and for this, we'll need to identify a couple of key points. So, the first key point is going to occur when X is equal to -7, since that is the X value of our lower bound. So we'll need to calculate F of -7, and this is going to be equal to the absolute value of the quantity of -7, minus 1 in quantity, plus the absolute value of the quantity of minus 7 plus 5 in quantity. And when we evaluate this expression, this is going to be equal to 10. The next point that we want to look at is when X is equal to -5, because that's going to make the quantity of X + 5 equal to 0, which shows up in our second absolute value function in F of X. So we need to calculate F of minus 5. And this is going to be equal to the absolute value of the quantity of -5, minus 1 in quantity, plus the absolute value of the quantity of minus 5, plus 5 in quantity. And when we evaluate this expression, this is going to be equal to 6. The next value of X that we want to look at is when X is equal to 1, and that's because the quantity of X minus 1 will be equal to 0, which is the quantity that shows up in the first absolute value function in F of X. So we'll need to calculate f of 1, and that's going to be equal to the absolute value of the quantity of 1 minus 1 in quantity, plus the absolute value of the quantity of 1 + 5 in quantity, and this is going to be equal to 6 when we evaluate that expression. And the last point that we want to look at is when X is equal to 5, which is the upper bound of our interval. So we'll calculate F of 5. And this is going to be equal to the absolute value of the quantity of 5 minus 1 in quantity, plus the absolute value of the quantity of 5 plus 5 in quantity. And when we evaluate this expression, this is going to be equal to 14. So now we just want to plot these points on a graph. So the first point is going to occur when X is equal to -7 and Y is going to be equal to 10. The next point is going to occur when X is equal to -5, and Y is equal to 6. Next point is going to be when X is equal to 1 and Y is equal to 6. And for the final point, we'll have X equal to 5 and Y equal to 14. And since we're taking the sum of two absolute value functions, we just need to connect these points with straight lines to get our graph. So doing so, we'll get the graph that we have drawn on the screen. And now we can determine the absolute extreme values by inspection of our graph. So we'll see that we have an absolute maximum. Whenever a function F of X reaches its maximum value, and this occurs. At X equal to 5. So we have an absolute maximum of 14. At x equal to 5. And if we look at the absolute minimum, that's going to occur whenever F of X reaches its minimum value, and we actually have a minimum value of 6, and this is actually on the closed interval from -5 to 1. And so those are the answers that this problem was looking for. So I hope that this explanation has been useful, and I'll see you in the next video.

Theory and Examples

[Technology Exercise] Graph the functions in Exercises 63–66. Then find the extreme values of the function on the interval and say where they occur.

f(x) = |x − 2| + |x + 3|, −5 ≤ x ≤ 5

Key Concepts

Absolute Value Function

Piecewise Functions

Extreme Values on an Interval

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