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marishachu [46]
3 years ago
8

PLEASE HELP!!!! Which of these pair of functions are inverse functions?

Mathematics
1 answer:
Mamont248 [21]3 years ago
3 0

Answer:

Option B and C are correct.

Step-by-step explanation:

Inverse function: If both the domain and the range are R for a function f(x), and if f(x) has an inverse g(x) then:

f(g(x)) = g(f(x)) = x for every x∈R.

Let f(x) = \frac{1}{2}(\ln(\frac{x}{2}) -1) and g(x) = 2e^{2x+1}

Use logarithmic rules:

  • ln e^a = a
  • e^{lnx} = x
  • \ln a^b = b\ln a

then, by definition;

f(g(x)) = f(2e^{2x+1}) =\frac{1}{2}(\ln(\frac{2e^{2x+1}}{2})-1) = \frac{1}{2}(\ln(e^{2x+1}}){-1) = \frac{1}{2} (2x+1-1) =\frac{1}{2}(2x) = x

g(f(x)) = g(\frac{1}{2}(\ln(\frac{x}{2}) -1)) = 2e^{2({\frac{1}{2}(\ln(\frac{x}{2}) -1})+1 2e^{(\ln(\frac{x}{2}) -1+1}=2e^{\ln(\frac{x}{2})} =2\cdot \frac{x}{2} = x

Similarly;

for f(x) = \frac{4 \ln(x^2)}{e^2} and g(x) = e^{\frac{e^2 \cdot x}{8} }

then, by definition;

f(g(x)) = f(e^{\frac{e^2 \cdot x}{8}}) =\frac{4 \ln {(\frac{e^2 \cdot x}{8})^2}}{e^2} = \frac{8 \ln {(\frac{e^2 \cdot x}{8})}}{e^2} =\frac{8\frac{e^2\cdot x}{8} }{e^2}=\frac{8e^2 \cdot x}{8e^2}=x

Similarly,

g(f(x)) = x

Therefore, the only option B and C are correct. As the pairs of functions are inverse function.

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