{Use of Tech} A family of superexponential functions Let ƒ(x) ...

Question

{Use of Tech} A family of superexponential functions Let ƒ(x) = (a + x)ˣ , where a > 0.

b. Describe the end behavior of f (near the left boundary of its domain and as x→∞).

Accepted Answer

Welcome back everyone to another video. Let H of Z be equal to 22 + 1 raise the power of C, where C is a real number. What is the end behavior of H as C approaches the left boundary of its domain and as C approaches infinity? So we're given our exponential function H of C equals 2 Z + 1, raise the power of C. And we can identify its domain understanding that our base must be positive, right? This is very important for. Exponential functions in the form of F of X raised to X, right? In a general case, even if we take a raise the power of X, we have to report that A must be positive. So what we have to do is simply set. 2 Z + 1. Being greater than 0, so if we rearrange, we're getting 2 z greater than -1 and Z must be greater than -12, meaning the domain of this function is Z. Belongs to the interval from -12 up to infinity. So our left boundary is -12, and we want to check what happens as he approaches -12 from the right. So basically what we have to do is check our end behavior as C approaches -15 from the right. We want to understand what happens to H of Z. And we're going to perform some analysis, so. If we analyze 2 Z + 1. And Z approaches -12 from the right. If we substitute -12, we get. -1 + 1, that's 0, but we are approaching 0 from the right, correct? That's basically because our negative term is slightly greater than -1, and if we add 1, we get 0 from the right. So 2 Z + 1 approaches 0 from the right and C approaches negative 1/2 from the right. And that means The natural log, what we're going to do here is basically take the natural log off that valley. LN of 2 Z. Plus one Approaches Negative infinity, right, if we recall the graph of a logarithm when we approach here from the right, our logarithm tends to negative infinity. Now why are we taking natural logarithm? Well, basically it will simplify the analysis. Notice that if we Use one of the properties. the power of L and of H is simply equal to H, right? So. We will be able to use this property along with the power rule of logarithms to bring down the exponency, right? So we have these two facts and then we can say that LN. Of 2 Z + 1. Raise the power of C. Which is Z multiplied by LN of 2 Z + 1. This is where we have applied the Power rule. Now with this term, Approaches -1.5 multiplied by negative infinity. Which basically gives us positive infinity. And now knowing this. We can essentially understand the behavior of HOC. So H of C. Which is equal to E. To the power of Z. LN of 2 Z + 1. Approaches to the power of infinity which is infinity. This is why. The usage of natural logarithms simplifies the problem. We can see that as Z approaches -15 from the right, H of Z approaches infinity. And now for the second case, we want to check the end behavior as Z approaches infinity. So as C. Approaches infinity. HOC It's going to approach. The value that we are going to evaluate. So in this case, we once again want to check H of C, which is equal to E to the power of Z LN of 2 Z + 1. Now if C approaches infinity, we get infinity multiplied by Llan of infinity, which is infinity. So basically e to the power of infinity is infinity once again. So and behavior as he approaches infinity would be H of C approaching infinity, and that's it, we have the complete analysis for this problem and our final answers. So as Z approaches -1 half from the right, H of Z approaches infinity and it approaches infinity as well as. As he approaches infinity. Thank you for watching.

{Use of Tech} A family of superexponential functions Let ƒ(x) = (a + x)ˣ , where a > 0.b. Describe the end behavior of f (near the left boundary of its domain and as x→∞).

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{Use of Tech} A family of superexponential functions Let ƒ(x) = (a + x)ˣ , where a > 0.

b. Describe the end behavior of f (near the left boundary of its domain and as x→∞).