Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 19–40:

a. Find the open intervals on which the function is increasing and those on which it is decreasing.

b. Identify the function’s local extreme values, if any, saying where they occur.

f(x) = (x² − 3) / (x − 2), x ≠ 2

Accepted Answer

Welcome back, everyone. In this problem for the function H of X equals 3 X2 minus 4 divided by X minus 1, where X is not equal to 1, find the intervals of increase and decrease and any local Xtrema. Now, in order for us to find um the, the intervals of increase and decrease in any local extra of for function, first, we'll need to find its critical points. And critical points occur where the first derivative equals 0 or where the first derivative is undefined. So let's differentiate F H X sorry, try to find those critical points and then determine which ones fit the criteria that we're searching for. So now we know that at the critical points, OK. The first derivative equals 0 or is undefined. So how can we differentiate H of X? Well, notice that it's a quotient, so we can differentiate it using the quotient rule. Recall that in differentiation, the quotient rule basically tells us that if we have a quotient Y, which equals u divided by V where U and V are functions of X, then the derivative of Y. Is going to be equal to Vu minus UV divided by V2. If we apply that to our function H of X, we can let you or numerator be equal to 3X2 minus 4. And our denominator V be equal to X minus 1. In that case, that means the derivative of U will be 6 X and the derivative of V will be equal to 1, OK? Which means then if we apply the quotient rule, the derivative of H of X is going to be equal to VU6X multiplied by X minus 1. Minus UV 1 multiplied by 3 x squared minus 4. All divided by v squared X minus 1 squared. Now, if we expand our terms here, this will be 6 X2 minus 6 X minus 3X2 + 4, all divided by X minus 1 squared. And when we simplify, that is going to be equal to, OK, uh 6 X. Sorry, 3 X squared. OK, it's gonna be equal to 3 x 2. Minus 6 x + 4 all divided by X minus 12. OK, so this is the derivative of H of X. No, this is what we want to figure out where it's either going to be equal to 0 or where it's going to be undefined. Well, for starters, we know that our first derivative is going to be undefined if our denominator is equal to zero. So are there any values that would make it equal to 0? Well, if X equals 1, it would be equal to 0, but remember, we were already told in our problem statement that X equals 1 is not in the domain. Thus, there are no values that are going to make our function undefined. What about values that would make it equal to 0? That is our first derivative equal to 0. Well, At the point where it equals 0, then our numerator would be equal to 0. In other words, 3 x 2 minus 6 X + 4 would be equal to 0. What x value would make this true? Well, we can solve for X using the quadratic formula. Now if we let A be 3, B negative 6, and C4. Then X is going to be equal to -6 plus R minus the square root of -6 squared. -4 multiplied by 3 multiplied by 4. All divided by 2 multiplied by 3 by the quadratic format. No, this is going to be equal to 6 plus or minus the square root of 36, 6 squared is 36 minus 48. That would be 4 multiplied by 3 multiplied by 4, and 36 minus 48 equals -12. So it's 6 plus R minus the square root of -12 divided by 6. But no notice here that the square root of a negative number will give us a complex or imaginary number. So because our discriminant is negative, there are no real solutions for X. Thus, the numerator 3X2 minus 6 X + 4 does not equal 0 for any real X, OK? So it's not equal to 0 for any real value of X. Now, what does that mean for us? Well, now that we know that it's neither undefined nor is it equal to zero for any real value of X, OK. Then We can take a good look at the sign of the numerator. They thus help us to try and figure out if it tells us something about whether or or whichever intervals or function will increase or decrease. No, like we said. Our quadratic has no real roots, and we already determined that it is always positive because the discriminant was negative and also the leading coefficient 3 is positive, so that the parael opens up and never crosses the x axis. So, the numerator, 3 X2 minus 6 X + 4 of our first derivative is always positive for all X that's not equal to 1. So now since. Or function or first derivative. Sorry, the numerator and the denominator of our first derivative are positive. That means that our first derivative is also going to be positive for all X that's not equal to 1. And since it's positive for all of those values, that means that the function is always increasing, OK? So thus our function H of X will be increasing. On the intervals where x is not equal to 1, so that is from negative infinity to 1, open and closed parentheses unioned with 1 to positive infinity opened and closed parentheses. OK. And no, because our first derivative is positive everywhere in the domain of H of X except where the function is not defined, then they are low local extrema. The function is always increasing in both intervals. Thus, H of X is increasing on the intervals open parentheses negative infinity 1 closed parentheses unioned with open parentheses 1, the infinity closed parentheses. Thanks a lot for watching and oh, sorry, before I forget, there are no local extrema. Let me make a note of that. There are no local extrema. I almost stopped the video, didn't I? No, we can stop it. Thanks for watching everyone. I hope this video helped.

Identifying Extrema

In Exercises 19–40:

a. Find the open intervals on which the function is increasing and those on which it is decreasing.

b. Identify the function’s local extreme values, if any, saying where they occur.

f(x) = (x² − 3) / (x − 2), x ≠ 2

Key Concepts

Derivative and Critical Points

Intervals of Increase and Decrease

Local Extrema

Watch next