Which statement identifies how to show that j(x) = 11.6ex and k(x) = In (StartFraction x Over 11.6 EndFraction) are inverse functions? All that needs to be shown is that j(k(x)) equals x. All that needs to be shown is that j(k(x)) equals 1. It must be shown that both j(k(x)) and k(j(x)) equal x. It must be shown that both j(k(x)) and k(j(x)) equal 1.

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Answer:

It must be shown that both j(k(x)) and k(j(x)) equal x

Step-by-step explanation:

Given the function  j(x) = 11.6[tex]e^x[/tex] and k(x) = [tex]ln \dfrac{x}{11.6}[/tex], to show that both equality functions are true, all we need to show is that both  j(k(x)) and k(j(x)) equal x,

For j(k(x));

j(k(x)) = j[(ln x/11.6)]

j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}

j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)

j[(ln x/11.6)] = 11.6 * x/11.6

j[(ln x/11.6)] = x

Hence j[k(x)] = x

Similarly for k[j(x)];

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

k[11.6e^x] = ln (11.6e^x/11.6)

k[11.6e^x]  = ln(e^x)

exponential function will cancel out the natural logarithm leaving x

k[11.6e^x]  = x

Hence k[j(x)] = x

From the calculations above, it can be seen that j[k(x)] =  k[j(x)]  = x, this shows that the functions j(x) = 11.6[tex]e^x[/tex] and k(x) = [tex]ln \dfrac{x}{11.6}[/tex] are inverse functions.

Answer:

the answer is C

Step-by-step explanation:

I did the test and got it right

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