Use the second derivative test to find the local extrema of the given function.f(x)=x2−4x2+1f\left(x\right)=\frac{x^2-4}{x^2+1}f(x)=x2+1x2−4

Local min of − 4 -4 − 4 at x = 0 x=0 x = 0

Use the second derivative test to find the local extrema of the given function. f ( x ) = x 2 − 4 x 2 + 1 f\left(x\right)=\frac{x^2-4}{x^2+1} f ( x ) = x 2 + 1 x 2 − 4

In this problem, we're asked to use the second derivative test to find the local extrema of the function. Now, in order to use the second derivative test, we need to first find where our first derivative is equal to 0. Once we find any values for which that is true, we can plug those values into our second derivative. The sign of our second derivative will tell us whether we have a local maximum or a local minimum. So let's start here by just finding that first derivative so we can find where it's equal to 0. So finding our first derivative here, since I have one function being divided by another function, I need to use the quotient rule. Remember our memory tool for the quotient rule, low d high minus high d low over the square of what's below. So my bottom function, x squared plus 1 times the derivative of that top function, which is 2x minus high d low, so x squared minus 4 times the derivative of that bottom function, which is again 2x, and then divided by the square of what's below, that's x squared plus 1 squared. And we can go ahead and distribute these 2 x's in here so that we can do some simplification. So this gives us 2x cubed plus 2x and then a minus here 2x cubed minus 8x. Now because this is subtracting, this makes this subtraction and this one addition. So this 2x cube cancels with our negative 2x cubed, and then we have 2x +8x. So that gives us 10x. All of this is still over that x squared plus 1 squared. So our first derivative here ends up being 8x or, I'm sorry, 10 x, adding the 8x and the 2x together over x squared plus 1 squared. Now we wanna find where this is equal to 0. So we can do that by taking our our numerator, setting that equal to 0. So 10x equals 0, that results in a critical point of x equals 0. Now here, because we are working with a rational function, we do need to think about if our denominator doesn't exist or if our denominator will equal 0, making this first derivative not exist. So if I have x squared plus 1, I set that equal to 0, that results in x squared is equal to negative one. If I were to take the square root on both sides, the square root of negative one is an imaginary number. Because this doesn't result in a real number, we don't have to worry about it and we can just proceed in using that one critical point that we found and now plugging it into that second derivative. So to find that second derivative, we wanna differentiate this first derivative. So f double prime of x, again, we have to use the quotient rule, low d high minus high d low for that numerator. So I have x squared plus 1 squared times 10, that's the derivative of the top, minus high d low, so 10 x times the derivative of that bottom function that gives us 2 times x squared plus 1 times 2 x having to use the chain rule there. All of this over the square of what's below because what's below is already squared, squaring it again will make it to the power of 4. So this is x squared plus 1 to the power of 4. Now this looks pretty messy, but we can go ahead and simplify a lot because these all have a common term of x squared plus 1. So I can cancel one of those, so this x squared plus 1 will go away. This squared one will just become x squared plus 1 not squared, and then this 4 on the bottom becomes a 3 having canceled just one of those terms. Now if I simplify this a little bit more, I wanna distribute that 10 into my parenthesis here. So this gives me 10 x squared plus 10 and then minus 10 x times 2 times 2 x. 10x times 2 gives me 20x times 2x gives me 40x squared. Then this is divided by x squared plus 1 cubed. Now we can do a final bit of simplification combining our like terms. This gives us negative 30 x squared plus 10, all divided by x squared plus 1 cubed. Okay. So we have our fully simplified second derivative. Remember, this is where we wanna plug our values that we found in step 1 into this second derivative and pay attention to the sign. So plugging 0 in here, remember, that's the value we're working with. So if I go ahead and plug 0 into my second derivative and find f double prime of 0, this is negative 30 times 0 squared plus 10 over 0 squared plus 1 cubed. Now remember, we only really care about the sign of this value. Because this negative 30 times 0 squared cancels, I'm left with a positive 10 on the top. Then on the bottom, 0 squared plus 1 cubed, this will result in a positive value as well. So we know that this second derivative is going to be positive when x is equal to 0. Because of that, that tells us that this point at x equals 0 is a local minimum. So based on what we found here, we know that we have a local minimum at x equals 0. But what actually is this value? We can find the value of this minimum here by plugging that critical point x equals 0 back into our original function. Remember, our original function here was x squared minus 4 over x squared plus 1. So we can plug that 0 back in to get this minimum value, f of a 0 here. So this is equal to 0 squared minus 4 over 0 squared plus 1. This gives us negative 4 over 1, which we know is just negative 4. So So this local minimum value, we know that it is located at 0 negative 4. We could also state this as we have a local minimum value of negative 4 at x equals 0. And that's all for this one. Let me know if you have questions, and I'll see you in the next one.

Use the second derivative test to find the local extrema of the given function. f ( x ) = x 2 − 4 x 2 + 1 f\left(x\right)=\frac{x^2-4}{x^2+1} f ( x ) = x 2 + 1 x 2 − 4

Local min of − 4 -4 − 4 at x = 0 x=0 x = 0

Local max of 10 10 10 at x = 0 x=0 x = 0

Local min of − 4 -4 − 4 at x = − 2 x=-2 x = − 2 and Local max of 10 10 10 at x = 2 x=2 x = 2

5. Graphical Applications of Derivatives

The Second Derivative Test

Calculus

5. Graphical Applications of Derivatives

The Second Derivative Test

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Multiple Choice

Use the second derivative test to find the local extrema of the given function.
$f\left(x\right)=\frac{x^2-4}{x^2+1}$

Local max of $10$ at $x=0$

Local min of $-4$ at $x=0$

Local min of $-4$ at $x=-2$ and Local max of $10$ at $x=2$

No local extrema

Verified step by step guidance

First, find the first derivative of the function $ f(x) = \frac{x^2 - 4}{x^2 + 1} $ using the quotient rule. The quotient rule states that if you have a function $ \frac{u}{v} $, its derivative is $ \frac{u'v - uv'}{v^2} $.

Apply the quotient rule: Let $ u = x^2 - 4 $ and $ v = x^2 + 1 $. Then, $ u' = 2x $ and $ v' = 2x $. Substitute these into the quotient rule formula to find $ f'(x) $.

Simplify the expression for $ f'(x) $ and set it equal to zero to find the critical points. Solve the resulting equation to find the values of $ x $ where the first derivative is zero.

Next, find the second derivative $ f''(x) $ by differentiating $ f'(x) $ again. Use the quotient rule or product rule as necessary to find $ f''(x) $.

Evaluate $ f''(x) $ at each critical point found in step 3. Use the second derivative test: if $ f''(x) > 0 $, the function has a local minimum at that point; if $ f''(x) < 0 $, the function has a local maximum. If $ f''(x) = 0 $, the test is inconclusive.

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