Encontre os valores máximo e minimo absolutos de f no intervalo dado. f(x)=x^-2 lnx, [(1)/(2),

Encontre os valores máximo e minimo absolutos de f no...

Key Concepts

Extreme Value Theorem

This theorem states that if a function is continuous on a closed interval, then it is guaranteed to attain both a maximum and a minimum value on that interval. This concept is essential when searching for absolute extrema, as it assures that the process of evaluating the function at critical and endpoint values will cover all possible extrema.

Critical Points

Critical points are points in the interior of the interval where the derivative of the function is either zero or undefined, meaning the function potentially changes behavior from increasing to decreasing (or vice versa). These points are important candidates for local extrema, and in the context of finding absolute extrema, they must be examined along with the interval’s endpoints.

Derivative Calculation

To identify critical points and analyze the behavior of a function, calculating the derivative is a crucial step. The derivative provides information about the function's rate of change, allowing us to solve for where the derivative is zero or does not exist, which yields potential locations for local maxima and minima.

Endpoint Evaluation

Because the Extreme Value Theorem ensures that absolute extrema occur on a closed interval, it is mandatory to evaluate the function at the endpoints of the interval. These endpoint values, together with the function values at critical points, determine the absolute maximum and minimum on the interval.

Transcript

00:01 Let's find the absolute minimum and absolute maximum values of the function f of x equal x to negative 2 times natural logarithm of x on the closed interval from one half to four.

00:17 So first we know that this function is continuous on the given interval because the point x equals zero is not there and in fact the natural algorithms are defined for positive values of x.

00:32 But here we have an interval which is completely on the positive values of x.

00:41 So the functions continue there, and because the interval is close, we know that the function attains its extreme values on that interval, and the points where the extreme values of f are attained, it can be either the endpoints of the interval or the critical, critical numbers of f.

01:07 So we start by calculating the first derivative of f.

01:13 And that's a product here.

01:15 We have x squared x to a negative two times, natural logarithm of x.

01:20 We can see it also as a fraction, as a quotient, like natural algorithm of x over x square.

01:28 But let's see it like a product.

01:31 And then we apply the product rule for differentiation.

01:34 So we get the derivative of x to a negative 2 is negative 2 x to the negative 3 times the natural logarithm of x plus x to the negative 2 times the derivative of natural logarithm of x which is 1 over x.

01:53 So we get negative 2 natural logarithm of x over x is the first term here plus 1 over xq.

02:05 And that becomes 1 minus 2 natural of x over x cubed.

02:14 So that's the derivative and this derivative is not defined as zero.

02:21 That does not exist at zero and in fact is also defined only for positive values of x.

02:28 So zero is not a critical number of f because even though the derivative does not exist there, that value 0 is not in the domain of the function.

02:45 So we can say that f derivative at 0 does not exist, but 0 is not in the domain of f, so it is not a critical number of f.

03:31 So we got to solve the equation.

03:36 First derivative of f equals 0, and that's the same as the equation 1 minus 2, natural logarithm of x over x cubed.

03:44 And we are supposing on these equivalences here that x is positive.

03:54 So that we can take the natural logarithm.

03:57 So this is equivalent to the equation 1 minus 2, natural logarithm of x equal to 0, which is equivalent to natural logarithm of x equal to 1⁄2.

04:11 And because the exponential function is injective function, we get that for the positive numbers, we get that this equivalent to e to the natural logarithm of x equal to e to the one -half, and this equivalent to, because the natural algorithm and the exponential function are inverse to each other, we get x equal e to the one -half, which is squared of e.

04:40 And if we calculate that, that is about square root of e is about 1 .6487, 187, 1 2 ,070.

05:16 Okay, this is, so this is the only x equal square root of e is the only critical number.

05:40 Let's see in fact that this number is in the interval one -half's four because one -half is 0 .5 so between 0 .5 and 4 is the number 1 .6487.

05:54 So it's the other critical number of f in the interval 1 half of 4.

06:06 In fact, this is the only critical number of function in its extended domain which is a positive numbers...

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