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:
for every x∈R.
Let
and ![g(x) = 2e^{2x+1}](https://tex.z-dn.net/?f=g%28x%29%20%3D%202e%5E%7B2x%2B1%7D)
Use logarithmic rules:
then, by definition;
= ![\frac{1}{2}(\ln(e^{2x+1}}){-1) = \frac{1}{2} (2x+1-1) =\frac{1}{2}(2x) = x](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Cln%28e%5E%7B2x%2B1%7D%7D%29%7B-1%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%282x%2B1-1%29%20%3D%5Cfrac%7B1%7D%7B2%7D%282x%29%20%3D%20x)
![2e^{(\ln(\frac{x}{2}) -1+1}=2e^{\ln(\frac{x}{2})} =2\cdot \frac{x}{2} = x](https://tex.z-dn.net/?f=2e%5E%7B%28%5Cln%28%5Cfrac%7Bx%7D%7B2%7D%29%20-1%2B1%7D%3D2e%5E%7B%5Cln%28%5Cfrac%7Bx%7D%7B2%7D%29%7D%20%3D2%5Ccdot%20%5Cfrac%7Bx%7D%7B2%7D%20%3D%20x)
Similarly;
for
and ![g(x) = e^{\frac{e^2 \cdot x}{8} }](https://tex.z-dn.net/?f=g%28x%29%20%3D%20e%5E%7B%5Cfrac%7Be%5E2%20%5Ccdot%20x%7D%7B8%7D%20%7D)
then, by definition;
= ![\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](https://tex.z-dn.net/?f=%5Cfrac%7B8%20%5Cln%20%7B%28%5Cfrac%7Be%5E2%20%5Ccdot%20x%7D%7B8%7D%29%7D%7D%7Be%5E2%7D%20%3D%5Cfrac%7B8%5Cfrac%7Be%5E2%5Ccdot%20x%7D%7B8%7D%20%7D%7Be%5E2%7D%3D%5Cfrac%7B8e%5E2%20%5Ccdot%20x%7D%7B8e%5E2%7D%3Dx)
Similarly,
g(f(x)) = x
Therefore, the only option B and C are correct. As the pairs of functions are inverse function.
(3,-1)(4,3)
slope(m) = (3 - (-1) / (4 - 3) = 4/1 = 4
y = mx + b
slope(m) = 4
use either of ur sets of points....(4,3)...x = 4 and y = 3
now sub and find b, the y int
3 = 4(4) + b
3 = 16 + b
3 - 16 = b
-13 = b
so ur equation is : y= 4x - 13 <==
Answer:
Ok
Step-by-step explanation:
Step-by-step explanation:
hope this helps........