Find the absolute maximum and absolute minimum values of f on the given interval. f(x)=x+(1)/(x)

Find the absolute maximum and absolute minimum values of f...

Key Concepts

Optimization on Closed Intervals

This concept involves the process of finding the extreme values (maximum and minimum) of a function when it is restricted to a closed interval. The strategy typically includes finding critical points by setting the derivative equal to zero, evaluating the function at these points, and then comparing these values with those at the interval’s endpoints to identify the absolute extrema.

Extreme Value Theorem

This theorem states that if a function is continuous on a closed interval, then it must attain both a maximum and a minimum value on that interval. It is a fundamental result in calculus that guarantees the existence of absolute extrema under the proper conditions, which is crucial when optimizing functions over specific domains.

Critical Points

Critical points are values in the domain of a function where its derivative is zero or undefined. These points are significant because they can be potential candidates for local maximum or minimum values. Identifying critical points is a key step in determining the behavior of a function within an interval during optimization problems.

Derivative Calculation

Calculating the derivative of a function allows us to understand its rate of change and identify critical points. The derivative is used in optimization to find where the function’s increasing or decreasing behavior changes, which helps in locating potential extrema.

Endpoint Evaluation

In optimization problems over closed intervals, it is essential to consider the values of the function at the endpoints of the interval. This is because absolute extrema could occur at these boundaries, and neglecting them might lead to incomplete analysis of the function's behavior.

Transcript

00:01 Here we're looking for the extreme values of this function f of x equals x plus 1 over x on the closed interval from 0 .2 to 4.

00:11 So before i do anything, i want to find the domain of this function.

00:15 Notice that i have an x -endominator here, and i can't divide by 0, so therefore we have to exclude 0 from the domain.

00:24 Now we're going to find the first derivative f prime of x.

00:28 Before i do that, i'm going to rewrite the original function and just bring this xx up to the numerator, so i have x plus x to negative 1 instead of x plus 1 over x.

00:38 I'm doing this so i can use the power rule to differentiate.

00:43 So using the power rule, the derivative of x is just 1, and the derivative of x to the minus 1, i have to multiply down by this exponent and then subtract 1 from the exponent.

00:54 So i get the derivative f prime of x equals 1 minus x to the minus 2.

01:00 I'm going to rewrite the derivative here, just so it's easier for, for us to find the critical values.

01:06 I'm going to bring this negative exponent down into the denominator, so it becomes a positive exponent.

01:13 And then we're going to get a common denominator so i can rewrite the derivative as one big fraction.

01:18 So i'm going to multiply here by x squared over x squared, and that gives us one fraction x squared minus one all over x squared.

01:28 Now to find the critical values, there are two things i need to check.

01:32 The first is where the derivative equals zero.

01:34 So since we have a fraction, the only time a fraction equals zero is when the numerator equals zero...

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