Find the inverse function for: f(x) = 6(ex-1) + 2 Then verify that the functions are inverses algebraically (show that f(g(x)) = x and g(f(x)) = x).

Answer 1

The first thing to do for this is to replace f(x) with y. This is then y=6(e^x-1)+2.

Then, we swqp x and y to get x=6(e^y-1)+2

Now, we solve for y, and the first step of that is to subtract 2 on both sides: x-2=6(e^y-1)

Then, divide both sides by 6, to get (x-2)/6=e^y-1

Next, add 1 to both sides, so e^y=1+(x-2)/6

Now comes the difficult part. To remove y from the exponent, we need to do ln on both sides, so we have ln(e^y)=ln(1+(x-2)/6)

On the left side, the exponent rule makes it into y*ln(e), but ln(e)=1, so this is now y=ln(1+(x-2)/6), or g(x)=ln(1+(x-2)/6)

Now, to confirm that they are, indeed, inverses. We do g(f(x))=g(6(e^x-1)+2), and that is ln(1+((6(e^x-1)+2)-2)/6), and we need to simplify that ln.

Si, start by distributing the 6 and simplifying the fraction: ln(1+(6e^x-6+2-2)/6) or ln(1+(6e^x-6)/6)

Division makes this ln(1+e^x-1), which is ln(e^x)

Just like earlier, ln(e^x)=x, so g(f(x))=x

Now to verify f(g(x))=x: f(ln(1+(x-2)/6)) which is 6(e^{ln(1+(x-2)/6)}-1)+2

Just like ln(e)=1, e^ln is also 1, so this becomes 6((1+(x-2)/6)-1)+2

The 1's cancel, and this is 6((x-2)/6)+2

Then, the 6's also cancel since the first one is (technically) in the numerator and the second is in the denominatro

So, we have x-2+2, or x

Thus, f(g(x))=x, also

Thus, f(x)=6(e^x-1)+2 and g(x)=ln(1+(x-2)/6) are inverses of one another.

Answer 2

Let g(x) be the inverse function of f(x) and let f(x) = y so that f^-1(y) = x will be the inverse function.

Given: f(x)= 6(e^x-1)+2 = y

Solve for x in terms of y:

6(e^x-1) = y-2 (Subtracting 2 from both sides)

e^x-1 = (y-2)/6 (Dividing by 6 on both sides)

e^x = (y-2)/6 + 1 (Adding 1 on both sides)

x = ln{1+(y-2)/6}. (Taking logarithm on both sides)

Therefore, the inverse function is g(x) = ln{1+(x-2)/6}

To test algebraically,

consider f(g(x)) = f(ln{1+(x-2)/6})

= 6(e^{ln{1+(x-2)/6}}-1)+2

= 6 (1+(x-2)/6}-1)+2

= 6 (x-2)/6 +2 (1-1=0)

= x-2 + 2 (6/6=1)

= x (-2+2=0)

f(g(x)) = x

Similarly consider g(f(x))

= g(6(e^x-1)+2)

= ln{1+(6(e^x-1)+2-2)/6}

= ln{1+(6(e^x-1)/6} (-2+2=0)

= ln{1+(e^x-1)} (6/6=1)

= ln(e^x) (1-1=0)

= x

g(f(x)) = x

f(g(x)) = g(f(x)) = x

Therefore, f and g are inverse functions of each other.