Function defined by an integral Let ƒ(𝓍) = ∫...

Question

Function defined by an integral Let ƒ(𝓍) = ∫₀ˣ (t ― 1)¹⁵ (t―2)⁹ dt .

(c) For what values of 𝓍 does ƒ have local minima? Local maxima?

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says let H of X be equal to the integral from 0 to X of the quantity of T minus 4 and quantity rates to 11, multiplied by the quantity of T minus 5 and quantity rates to 6 dT. Determine the values of X where the function H of X has local minima or local maxima. So we need to find the values of X where our function H of X has either a local minima or a local maxima. So in order to do that, we first need to take a derivative of a function H of X with respect to X. We'll be calculating H of X. And here we'll be taking the derivative of an integral, where the variable that we're taking a derivative of is the upper limit of our limits of integration. And since the lower limit in our case is a constant, we can use the fundamental theorem of calculus to find this derivative. And so using that, we'll see that H of X is going to be equal to the quantity of X minus 4 in quantity raised to the 11th, multiplied by the quantity of X minus 5. In quantity rates to the 6th. So here we just replace T with X in our integram. So now we need to find a critical points, and recall that our critical points occur whenever our first derivative is either undefined or equal to zero. And H X is a polynomial, so it's defined everywhere. So the critical points are going only going to come from setting H X equal to 0. And so doing that, we'll have the quantity of X minus 4 in quantity raised to the 11th, multiplied by the quantity of X minus 5, in quantity raised to the 6th, equal to 0. So what we want to do is to solve this equation for X, and we can do that by setting each one of these factors equal to 0. So for the first factor we'll have the quantity of X minus 4 and quantity rates to 11 equal to 0. Taking the 11th root of both sides of that equation, we'll get X minus 4 is equal to 0, and solving for X by adding 4 to both sides, we'll get that X is equal to 4. So that is one of our critical points. We'll do the same thing for the other factor. So we'll have the quantity of X minus 5, in quantity rates to the 6th equal to 0. Taking the 6th root of both sides, we'll have X minus 5. is equal to 0, and adding 5 to both sides, the soft X will get that X is equal to 5. So these are two critical points. So now we want to use the first derivative test to see whether these critical points are local minima, or local maxima. So that means we need to look at the sign of H X. In the intervals that are bound by a plus minus infinity as well as these critical points. So, the first interval that we want to look at is going to be the open interval from minus infinity to 4. And to check the sign of HX on this interval, we'll just choose a value of X and evaluate H X. So we're going to choose that X is equal to 3, and so that means that H of 3 is going to be equal to the quantity of 3 minus 4. In quantity, raised to the 11th. Multiplied by the quantity of 3 minus 5 in quantity, raised to the 6. And when we evaluate this expression, we see that it is going to be equal to -64, which is less than 0. So, H of X is negative on this interval. So the next interval that we need to look at is going to be the open interval from 4 to 5. And so again, we're gonna choose a value of X on this interval and value weight H X. So we're going to choose that X going to be equal to 4.5. That means that each prime. Of 4.5. is going to be equal to the quantity of 4.5 minus 4 in quantity raised to the 11th, multiplied by the quantity of 4.5. -5 in 20 rates to the 6th. And if we simplify this expression, we see that this is going to be equal to. For the first term, we're gonna have 0.5. Raised to the power 11, and then that gets multiplied by minus 0.5. Raised to the 6th However, since we're raising a negative value to an even power, that means we'll have -1 to the 6th, which is just going to be equal to 1. So this is also equal to the quantity of 0.5 raised to the 11th, multiplied by the quantity of 0.5, raised to the power of 6, and both of those are positive quantities, so when we multiply them together, we will get an overall positive value. That means that this Um, value is going to be greater than 0, and so H X is going to be positive on this interval. And the final interval that we need to look at is going to be the open interval from 5 to infinity. And again, we'll choose a value of X, and so we'll choose X equal to 6, and evaluate H of X. So we'll evaluate H of 6. Which is going to be equal to. The quantity of 6 minus 4 in quantity raised to the power 11. Multiplied by the quantity of 6 minus 5, in quantity raised to the power of 6, which when we simplify this expression is going to be equal to 2, raised to the power 11, multiplied by 1, raised to the power 6. Which is just going to be equal to 2 raise to the power 11, which again, is a positive quantity. So, H 6 is greater than 0, so that means that H of X is going to be positive on this interval. So now we can use how these signs change on these different intervals to determine whether we have a local minimum or a local maximum at each of our critical points. So if we look at X equal to 4, the sign of H X is going from negative to positive, that means that we have a local minimum. And if we look at X equals to 5, we see that H of X is positive on both sides of that critical point. So, X equal to 5 is neither a local minimum nor a local maximum. So, that means that if we look at our final answer, that we're going to have A local minimum. At X equal to 4. And we're going to have a local maximum. At none of the values of X. So the local maximum is going to be none. And so those are the answers that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Function defined by an integral Let ƒ(𝓍) = ∫₀ˣ (t ― 1)¹⁵ (t―2)⁹ dt .
(c) For what values of 𝓍 does ƒ have local minima? Local maxima?

Key Concepts

Fundamental Theorem of Calculus

Critical Points

Second Derivative Test

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