Find the critical points of the given function. f ( t ) = 6 t t 2 + 1 f(t)=\frac{6t}{t^2+1} f ( t ) = t 2 + 1 6 t

Here we're asked to find the critical points of the given function and the function we're given here is f of t equals 6 t over t squared plus 1. Now remember that our critical points are going to be where our derivatives, so in this case, f prime of t, is either equal to 0 or it does not exist. So we need to start here by finding the derivative of our function. Now because here we're working with a quotient, we have one function being divided by another function, that means we need to use the quotient rule. Now we can remember our nice memory tool for the quotient rule. Low d high, so I'm gonna take that low that function on the bottom, t squared plus 1, multiply it by the derivative of that top function, which in this case is 6, then subtract high d low, so that's 6 t times the derivative of that bottom function, which in this case is going to be 2 t, then over the square of what's below, so taking my denominator of my original function t squared plus 1, and squaring it. Now we can do a bit of simplification here. We can distribute that 6 and we can also multiply this 6 t times 2 t. So this will give us 6 t squared plus 6 and then minus 12 t squared. Now again, we can do some simplification here because we have some like terms in that numerator. We have 6 t squared and negative 12 t squared. Now this will give us negative 6 t squared plus 6 and then that's all over t squared plus 1. That denominator has not changed. And then finally, one last step here, we can factor out a 6 from that numerator. So if we factor out that 6, we do a little bit of rearranging here just to make it look a bit nicer. That's going to be 1 minus t squared having factored that out and then this all gets divided by t squared plus 1 squared. Don't forget that squared on the bottom for your derivative. It's easy to forget that there. Now that we're here, we have our derivative. We have some more steps to take because we're looking for where our derivative is equal to 0 or where it does not exist. Now for these rational functions, when we want to find where our derivative is equal to 0, that means that we really just want to take our numerator and set it equal to 0. Now for our numerator to be equal to 0, that 6 doesn't really affect anything from this standpoint. And we can just take that 1 minus t squared and set that equal to 0. Now if I add t squared on both sides, that ends up giving me 1 equals t squared. If I take the square root, then that gives me my answer here. The square of a one is going to be plus or minus 1. So this gives me 2 critical points where t is equal to positive 1 and t is equal to negative 1. Now one final thing to think about here, is there anywhere where our derivative does not exist? Now we could go ahead and look at our derivative and figure that out, but something that I do want you to notice is if that if we go ahead and look at our original function and if we look at our derivative, they have really similar things in the denominator. In fact, they have the same exact function, t squared plus 1, except in our original function, it is just by itself, And in our derivative, it is squared. Now because these are the same exact things, that means that they're not going to exist at the same exact point. And our we will only have a critical point there if that point exists for our original function. So because our derivative and our original function don't exist at the exact same spot, we don't even have to worry about that. We already know that these are our only 2 critical points where t equals positive 1 and t equals negative 1. Let us know if you have any questions here, and let's keep going.

5. Graphical Applications of Derivatives

Finding Global Extrema

Calculus

5. Graphical Applications of Derivatives

Finding Global Extrema

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Multiple Choice

Find the critical points of the given function.
$f(t)=$\frac{6t}{t^2+1}$$

$t=0,t=6$

$t=1$

$t=-1,t=1$

No critical points

Verified step by step guidance

To find the critical points of the function $ f(t) = \frac{6t}{t^2+1} $, we first need to find its derivative $ f'(t) $.

Apply the quotient rule for differentiation, which states that if $ f(t) = \frac{u(t)}{v(t)} $, then $ f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} $. Here, $ u(t) = 6t $ and $ v(t) = t^2 + 1 $.

Calculate $ u'(t) = 6 $ and $ v'(t) = 2t $. Substitute these into the quotient rule formula to find $ f'(t) = \frac{6(t^2 + 1) - 6t(2t)}{(t^2 + 1)^2} $.

Simplify the expression for $ f'(t) $ to get $ f'(t) = \frac{6 - 6t^2}{(t^2 + 1)^2} $.

Set $ f'(t) = 0 $ to find the critical points. This gives the equation $ 6 - 6t^2 = 0 $. Solve for $ t $ to find the critical points, which are $ t = -1 $ and $ t = 1 $.

3:52m

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