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QuestionFind the inverse f−1(x)f^{-1}(x) of f(x)=e2x−3−2f(x)=e^{2x-3}-2 and its domain/range. Sketch ff and f−1f^{-1}. For g(x)=ln⁡3xg(x)=\ln 3x, find (f⋅g)(x)(f \cdot g)(x).

Studdy Solution

STEP 1

Assumptions1. The function is given as f(x)=ex−3−f(x)=e^{x-3}-. . We need to find the inverse of the function, f−1(x)f^{-1}(x), and state its domain and range.
3. We need to sketch the graphs of ff and f−1f^{-1} on the same coordinate axes.
4. We have another function g(x)=ln⁡3xg(x)=\ln3x and we need to find the composition of the functions, (f⋅g)(x)(f \cdot g)(x).

STEP 2

To find the inverse of a function, we first replace f(x)f(x) with yy.
y=e2x−−2y = e^{2x-}-2

STEP 3

Next, we swap xx and yy to find the inverse function.
x=e2y−3−2x = e^{2y-3}-2

STEP 4

We then solve this equation for yy to get the inverse function.
First, isolate the exponential term by adding2 to both sides.
x+2=e2y−3x +2 = e^{2y-3}

STEP 5

Next, take the natural logarithm (ln) of both sides to get rid of the exponential term.
ln⁡(x+2)=2y−3\ln(x +2) =2y -3

STEP 6

Finally, solve for yy to get the inverse function.
f−1(x)=12(ln⁡(x+2)+3)f^{-1}(x) = \frac{1}{2}(\ln(x +2) +3)

STEP 7

The domain of f−1(x)f^{-1}(x) is the set of all real numbers greater than -2, because the logarithm is only defined for positive numbers.The range of f−1(x)f^{-1}(x) is the set of all real numbers, because the logarithm function can take any real number as its output.

STEP 8

To sketch the graphs of ff and f−1f^{-1}, we note that the graph of f−1f^{-1} is a reflection of the graph of ff in the line y=xy=x.The graph of ff is an exponential function shifted3 units to the right and2 units down.The graph of f−1f^{-1} is a logarithmic function shifted2 units to the left and3 units up.

STEP 9

We have another function g(x)=ln⁡3xg(x)=\ln3x. We need to find the composition of the functions, (f⋅g)(x)(f \cdot g)(x), which means we substitute g(x)g(x) into f(x)f(x).

STEP 10

Substitute g(x)g(x) into f(x)f(x) to get (f⋅g)(x)(f \cdot g)(x).
(f⋅g)(x)=f(g(x))=e2(ln⁡3x)−3−2(f \cdot g)(x) = f(g(x)) = e^{2(\ln3x)-3}-2

STEP 11

implify the expression using the property of logarithms aln⁡b=ln⁡baa \ln b = \ln b^a.
(f⋅g)(x)=eln⁡(3x)−3−(f \cdot g)(x) = e^{\ln (3x)^-3}-

STEP 12

implify further using the property of exponentials and logarithms eln⁡a=ae^{\ln a} = a.
(f⋅g)(x)=(x)2−2=9x2−2(f \cdot g)(x) = (x)^2 -2 =9x^2 -2So, (f⋅g)(x)=9x2−2(f \cdot g)(x) =9x^2 -2.

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