if ƒ(x) = 1 / (3x⁴ + 5) , it can be shown that ƒ'(x) = 12x³ / ...

Question

if ƒ(x) = 1 / (3x⁴ + 5) , it can be shown that ƒ'(x) = 12x³ / (3x⁴ + 5)² and ƒ"(x) = 180x² (x² + 1) (x + 1) (x - 1) / (3x⁴ + 5)³ . Use these functions to complete the following steps.

d. Identify the local extreme values and inflection points of ƒ .

Accepted Answer

Find the local extrema and inflection points of the function F of X. If F of X equals 1 divided by 6 X the fourth plus 10. F of X equals -6 X to the third divided by 3x the 4th plus 52. And F X equals 90 X 2. X the 4th minus 1, all divided by 3 X the 4th plus 5, raises to the 3rd. We have 4 possible answers, and we're searching for the local max, local min, and the inflection points. Now, to find our Macs or min, we first need to find the critical point. To find the critical point, we will set our F of X equals to 0. We have 6 X the 3rd. Divided by 3 X to the 4th plus 5 squared. Equals 0. Now, if we multiply the denominator over, we end up getting negative 6 X3 equals 0. Which results in us getting X equals 0. This is our critical point. Now, we'll use that in a moment to find our maxim min, but let's first find our inflection points. To find our inflection points, we set F of X equals 0. Which means we have 90 X squared multiplied by X to the 4th minus 1. Divided by 3 X to the 4th plus 5, all cubed equals 0. Now we can multiply the denominator to get 90 X squad. Multiplied by X the 4th minus 1 equals 0. We can split these up. And then solve for X. For our left equation we get x equals 0, and for the right equation. We get X the 4th equals 1, or X equals -1 or 1. So now we have our inflection points. Next, let's determine our local max, min, and our concavity. Now, Our function increases. Where F of X is greater than 0. Decreasing for F of X is less than 0. We will use this to determine our concavity. We know our critical point is at 0, so we will check the interval from negative infinity to 0, and the interval from 0 to infinity. Let's take an example point. We say X equals -1. Our output would be 3:30 seconds. That is greater than 0, so this is increasing. If we plug in a number larger than 0, say X equals 1. We end up getting F 1 equals -330 seconds. Which is less than 0. Making that interval decreasing. We can then see That the interval. Before the critical point, is increasing interval, after the critical point. is decreasing. Next, let's find the concavity. We will check the sign of our 2nd derivative. F double prime Of X If it is greater than 0, We have an interval, that's concave up. Less than 0. Means it's concave down. So let's plug in values on our intervals. We'll have negative infinity to -1. -1 to 0. 0 to 1 And one to infinity. And if we find some example points, let's say X equals -2. F F 2. As equals To about 5400. Divided by 14887. That's a positive value, which means that interval is concave up. If we did F of -0.5 for our next interval, We end up getting about 0.15, which is concave down. For our 3rd interval, You plug in 0.5. We end up getting about 0.15. Which is once again concave down. And finally, for our last interval. If we plug in X equals 2, We end up getting F double F2. Which is about. 5400. Divided by 148. 8 77 Which is a number. That is positive, making this concave up. Finally, we can see that our cavity change. At 0, doesn't happen. is concave down the 40 and after zero. This then tells us that we have A local pool Maximum And X equals 0. And we have inflection points. Working cavity does change at X equals -1 and X equals 1. If we look at our possible answers, We can then see The answer to our problem is answer A. Local max at X equals 0, and inflection points at -1 and 1. There is no local minimum. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

if ƒ(x) = 1 / (3x⁴ + 5) , it can be shown that ƒ'(x) = 12x³ / (3x⁴ + 5)² and ƒ"(x) = 180x² (x² + 1) (x + 1) (x - 1) / (3x⁴ + 5)³ . Use these functions to complete the following steps.d. Identify the local extreme values and inflection points of ƒ .

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if ƒ(x) = 1 / (3x⁴ + 5) , it can be shown that ƒ'(x) = 12x³ / (3x⁴ + 5)² and ƒ"(x) = 180x² (x² + 1) (x + 1) (x - 1) / (3x⁴ + 5)³ . Use these functions to complete the following steps.

d. Identify the local extreme values and inflection points of ƒ .