Let

Question

Let $g\left(x ight)=\begin{cases}5x-2 & ext{if }x<1\ a & ext{if }x=1\ ax^2+bx & ext{if }x>1\end{cases}$ .

Determine values of the constants $a$ and $b$ , if possible, for which $g$ is continuous at $x equals 1$ .

Accepted Answer

Welcome back, everyone. Consider the function F of X equals 3X minus 4 if X is less than 1, M if X is equal to 1, and MX2 + N X + 1 if X is greater than 1. Find the values of M and N that make F continuous at X equals 1. A says M equals 1 and N equals -2. B says they are -1 and 2. C 1 and -1 and D-1 and 1. Now what is the condition that has to be satisfied to make F continuous at X equals 1? Well, for F to be continuous at X equals 1, then that means the limit. OK, the limit as X approaches 1 from the left of F of X. Must be equal to the limit as x approaches one from the right or F of X, which is going to be equal to f of 1. So it must be equal to F1 for our function to be continuous at that point. Now notice here that if X equals 1, then M is going to be F1. So if we can evaluate the limit as X approaches 1 from the left, it's going to be equal to F1, which will give us the value of M, and then we can use that to solve for the value of N as X approaches 1 from the right. So let's go ahead and do that. Now, for X being less than 1, OK, then F of X is going to be equal to 3X minus 4 based on our piece function. So if we want to evaluate the limit as X approaches one from the left of FF X, that means we'll be evaluating the limit, OK. Of 3 X minus 4. And to find this limit, we can substitute 1 into our expression. So that's going to be 3 multiplied by 1 minus 4, which is going to be 3 minus 4, which equals -1, OK? Therefore, That tells us then that the limit as x approaches 1 from the left of F of X, which is going to be equal to F1 for it to be continuous, which thus must be equal to the value of m is going to be -1, because since F of 1 equals -1, then M must equal -1. Know that we have a value for M we can solve for N using the right hand limit. Of our function at X equals 1, OK, because no, OK, when, uh sorry, 4 X greater than 1, that is from the right, then the value of F of X we're going to be using or the expression we're going to be using is going to be MX2 + N X + 1. So that means That the limit, OK. As X approaches one from the right of FF X is going to be the limit, OK. Of the expression m x2 + N X + 1 and we know what the value of m is and we also know based on our definition of what it means for the function F to be continuous that this is going to be equal to F1, which is -1. Now if we substitute our value for M here and the substitute 1 into our expression for X, OK, then. This means we're going to get -1. Multiplied by 1 squared plus n multiplied by 1 plus 1 being equal to -1. No -1 multiplied by 1 squared is going to be -1, so that's -1 + N + 1 equals -1, 1 + 1 equals 0. So no, let me just clean this up here. No, this tells us. OK, that N must be equal to -1. Therefore, the values of M and N that make F continuous are -1. If we look back on our answer choices, that means D is the correct answer. Thanks for watching everyone. I hope this video helped.

Let $g\left(x\right)=\begin{cases}5x-2 & \text{if }x<1\\ a & \text{if }x=1\\ ax^2+bx & \text{if }x>1\end{cases}$ .

Determine values of the constants $a$ and $b$ , if possible, for which $g$ is continuous at $x equals 1$ .

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