Question

[Will mark as brainliest]
What is the inverse of the function below?

f(x) = 2^x + 6

A.) f^-1(x) = 6 + log2x
B.) f^-1(x) = log2(x + 6)
C.) f^-1(x) = 6log2x
D.) f^-1(x) = log2(x - 6)

Answer 1

To invert this function, let's solve the expression for x. We start with

$f(x) = y = 2^x+6$

If we subtract 6 from both sides, we have

$y-6 = 2^x$

Now we need to manipulate both sides so that the right hand side will consist only of x. To do so, we must perform the inverse function of $2^x$.

$2^x$ is an exponential function, because the variable is at the exponent (note the difference, with, for example, $x^2$, where the exponent is always 2). The inverse function of an exponential function is the logarithm, taken with the same base as the exponential function.

So, the inverse function of $f(x) = 2^x$ is $f^{-1}(x) = log_2(x)$. Let's apply this function to both sides to get

$log_2(y-6) = x$

Which means that we can rename the variables and state that the inverse function is

$f^{-1}(x) = log_2(x-6)$

i.e. option D.)

Answer 2

What is the inverse of the function below?

f(x) = 2^x + 6

1) Replace "f(x)" with "y" y = 2^x + 6

2) Interchange x and y: x=2^y + 6

3) Solve this for y: 2^y = x - 6 Take the log to the base 2 of both sides:

y = (log to the base 2 of) (x-6)

4) replace y with

-1

f (x)

We obtain:

-1

f (x) = (log to the base 2 of) (x-6)