Let f(x) = \ln(7x-2) and g(x)=e^{7x-2} a. Find domain of f(x) b. Find domain of g(x) c. Find...


Let {eq}f(x) = \ln(7x-2) {/eq} and {eq}g(x)=e^{7x-2} {/eq}

a. Find domain of {eq}f(x) {/eq}

b. Find domain of {eq}g(x) {/eq}

c. Find {eq}f(1)+g(0) {/eq}

Domain of Exponential and Logarithmic Functions

Since exponential and logarithmic functions are inverses of one another, their domains are related. Specifically, the domain of a standard exponential function is all real numbers, but the range is all positive numbers. A logarithmic function has these two reversed, so the domain is all positive numbers and the range is all real numbers.

Answer and Explanation:

a) We can only take a logarithm of a positive number. Thus, the domain of this function is whatever values of x allow the expression within the logarithm is positive. We can find this as follows.

{eq}7x-2 > 0\\ 7x > -2\\ x > -\frac{7}{2} {/eq}

Thus, the domain of this function is all real numbers larger than -3.5.

b) An exponential function doesn't have the same restrictions as a logarithm. We can take the exponential function of any real number, provided that the expression in the exponent is defined. As this is a linear expression, it is defined everywhere. Thus, the domain of this function is all real numbers.

c) Lastly, we've been asked to calculate the sum of two function values. We can find them by evaluating the individual functions at these values first.

{eq}f(1) = \ln (7(1) - 2) = \ln 5\\ g(0) = e^{7(0)-2} = e^{-2} {/eq}

We acn now add these two values together.

{eq}f(1) + g(0) = \ln 5 + e^{-2} \approx 1.7447732 {/eq}

Learn more about this topic:

Behavior of Exponential and Logarithmic Functions

from Precalculus: High School

Chapter 15 / Lesson 16

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