Locating critical points Find the critical points of the follo...

Question

Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.

ƒ(t) = t/ t² + 1

Accepted Answer

Welcome back, everyone. Determine the critical points of the function F of X equals X divided by X2 + 4. Whenever we want to identify the critical points of a function, we first will have to differentiate that function. So in this case, we want to apply the quotient rule, right, because we have a rational function. According to the quotient rule, we first will take the derivative of the top function, which is the derivative of X, multiply by the bottom function which is X2 + 4. We then subtract X, which is the top function multiplied by the derivative of the bottom function x2 + 4, and we have to divide that by the square of the bottom function which is X2 + 4. And now let's evaluate each derivative. The derivative of X is 1, so we got X2 + 4 minus X multiplied by the derivative of X2 2 X and the derivative of 4 is 0. Now we are going to rewrite the denominator. And we're going to combine the like terms. So first of all, we have X2 minus 2 X2. This gives us 4 minus X2 in the numerator, right, because we have 4. Plus X2 minus 2 X2, so 4 minus X2 overall divided by X2 + 4 squad. Let's recall that the critical points occur where the derivative is equal to 0 or it does not exist in the domain of F of X. Now F of X is defined for all real values because the denominator is never equal to 0. We have a square of a number plus 4, so it's always positive, meaning. We want to number one check when the derivative is equal to 0. The derivative also exists for any X because the denominator's number equals to 0. So we are only going to check when the derivative is equal to 0, the denominator, well, it's number equal to 0, so that means the derivative exists within all the domain. And now if F of X is equal to 0 for a rational function, this means that we are setting our numerator equal to 0. While the denominator is not equal to 0, we already know that. So we can factor it out. We have 2 squad minus X2 equals 0, so let's apply the difference of squares factorization, which gives us 2 minus X multiplied by 2 + X is equal to 0, and this gives us two critical X coordinates. X1 is equal to 2. X2 is equal to -2. Because we have a product of two factors, if that product is 0, any of the two factors can be equal to 0. So we have the X values. Now we need to evaluate the Y values. So Y1 is simply the value of the function at 2. We're going to substitute 2 for every X, so we get 2 divided by 22 + 4. So 2 divided by 8 gives us 1/4, and now the second Y coordinate that corresponds to X2. Is the value of the function at -2. So we got -2. Divided by -2 2 + 4. So in this case, -2 divided by 8, which is 14. Therefore, we have two critical points. The first one is 2. 1/4. And the second one is 2 1/4. And those will be our final answers. Thank you for watching.

Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.ƒ(t) = t/ t² + 1

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Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.

ƒ(t) = t/ t² + 1