Question

Which statement identifies how to show that j(x) = 11.6ex and k(x) = In (StartFraction x Over 11.6 EndFraction) are inverse functions?

Answer 1

Answer:

The answer is "It must be shown that both $j(k(x))\ and \ k(j(x))=x$"

Step-by-step explanation:

Given:

$j(x) = 11.6 e^{x} \\\\k(x) =\ln\frac{x}{11.6}$

To show that both are equal functions, and  show that both$j(k(x))\ and\ k(j(x)) =x,$

For $j(k(x));$  

$j(k(x)) = j[(\ln \frac{x}{11.6})]\\\\j[(\ln (\frac{x}{11.6})] = 11.6e^{\ln (\frac{x}{11.6})}\\\\j[(\ln \frac{x}{11.6})] = 11.6(\frac{x}{11.6})$(The natural logarithm is canceled by exponential function)  

$j[(\ln \frac{x}{11.6})] = 11.6 \times\frac{x}{11.6}\\\\j[(\ln \frac{x}{11.6})] = x\\\\j[k(x)] = x\\\\for\ \ k[j(x)]:\\\\k[j(x)] = k[11.6e^x]\\\\k[11.6e^x] = \ln (\frac{11.6e^x}{11.6})\\\\k[11.6e^x] = \ln(e^x)$

Its natural logarithm leaving x will nullify expanding universe.

$k[11.6e^x] = x\\\\k[j(x)] = x$

In the question, it is seen that$j[k(x)] = k[j(x)] = x$, shows that the functions $j(x) = 11.6 e^{x} \ and \ k(x) = \ln \frac{x}{11.6}$ is inverse functions.

Answer 2

Answer:

C

Step-by-step explanation:

Ed2021