For what values of a and b is
g(x) = { ax + 2b, x ≤ 0

Question

For what values of a and b is

g(x) = { ax + 2b, x ≤ 0

x² + 3a – b, 0 < x ≤ 2

3x – 5, x > 2

continuous at every x?

Accepted Answer

Welcome back, everyone. For the function FFX given by FFX equals 2 X + 4C if X is less than or equal to -1, X2 + 5 C minus D if X is greater than 1 but less than or equal to 2, and 4 X minus 7 if X is greater than 2, find the values of C and D that make it continuous everywhere. A says C is 0 and D is 3. B says they are 3 and 0 respectively. C-3 and 3, and D 0 and -3. Now how can we find these values of C and D that will make our function FFX continuous everywhere? Well, if we can ensure continuity at a few points, find the possible equations for uh C and D for those points, that is where it involves the part of the equation where X is less than or equal to 1 and less than or equal to 2, then we should be able to use those simultaneous equations to solve for the values of C and D. So let's start with the first part of our P size function, where X is less than the equal to 1. We can try to first ensure continuity at X equals -1. Now how do we ensure continuity? Well, for FOX. To be continuous. For FX to be continuous at X equals -1, then we need the limit as X approaches -1 from the left of FX. To be equal to the limit as X approaches -1 from the right of F of X to be equal to F of -1. So we have three possible equations here, 3 possibly expressions that we could come up with in terms of C and D, OK. Now let's first start with our left hand limit. Now, our left hand limit, OK, as X approaches -1 from the left of FF X is going to be equal to the limit, OK, as X approaches -1 from the left of 2 X plus 4 C. Because remember, F X equals 2 X + 4C if X is less than or equal to -1, and that is approaching from the left. In this case, by direct substitution, then this limit will be 2 multiplied by -1 + 4C, which equals -2 + 4C. Now if we take into account here. Or or or let me highlight that in red actually. If we take into account our right hand limit. Then the limit as X approaches -1 from the right. OK. It's gonna be equal to. The limit As X approaches -1 of X2 + 5 C minus D. That would be equal to -1 squared plus a 5 C minus D which equals 1 + 5 C minus D. Finally, for continuity. So what that means then, sorry, not since the left-hand limit must be equal to the right-hand limit for continuity, then this would imply. That -2 + 4C would be equal to 1 + 5 C minus D. and if we arrange our equation here, that means D would be equal to C + 3. OK? So we can call this equation 1, that is for F of X to be continuous at X equals -1. Now, let's choose another point. Let's ensure continuity at X equals 2, OK? Now, for FFX then. To be continuous. Or to be continuous at x equals 2, then the limit as x approaches 2 from the left of FX must be equal to the limit as x approaches 2 from the right of F of X, which must be equal to f of 2. Now let's take our limits. Starting from the left, we know then that the limit as X approaches 2 from the left of F of X is going to be equal to the limit as X approaches 2 of X2. OK. Plus 5 C minus D. Substituting 2, that's 2 squared plus 5 C minus D, which equals 4 + 5 C minus D, OK? Next, if we take our right hand limit. OK, as X approaches to. Then, OK, it's gonna be equal to the limit, OK, as X approaches to of 4 X minus 7. That equals 4 multiplied by 2 minus 7. That's 8 minus 7, which equals 1, OK? So the limit from the right equals 1. Thus, for continuity at X equals 2, then 4 + 5 C minus D. Must be equal. To 1, which means then that 5 C minus d must be equal to -3. No, we can call this equation 2. Now that we have two equations for C and D, we can solve for C and D. Remember we said D equals C + 3. OK. So substituting. Substituting the value of D that we have in equation one. D equals C + 3. Into equation 2 to solve for C, then we're going to have 5 C minus C + 3 equal to -3. Which means that we're gonna have 4 C equal to -3 minus 3, OK. Which, well, -3 + 3, which means that 4 C equals 0, thus C equals 0, OK? So, therefore, we know that C equals 0. Now we can substitute C equals 0 into equation 1. So when we do that, then D is gonna be, which is equal to C + 3 is gonna be equal to 0 + 3, which equals 3. Therefore, to ensure that F of X is continuous everywhere, C has to equal 0, C has equals 0, sorry, and the D has to be equal to 3. If we look back at our answer choices, that tells us A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

For what values of a and b is
g(x) = { ax + 2b, x ≤ 0
x² + 3a – b, 0 < x ≤ 2
3x – 5, x > 2
continuous at every x?

Key Concepts

Continuity of Functions

Piecewise Functions

Limit and Value Matching

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