Question

question 4 of 25
find a formula for an exponential function $f(x) = a \cdot b^x$ such that $f(1) = 12$ and $f(3) = 108$.
\\(\bigcirc\\) $f(x) = \frac{4^x}{12}$
\\(\bigcirc\\) $f(x) = 12 \cdot 3^x$
\\(\bigcirc\\) $f(x) = 49x^{-37}$
\\(\bigcirc\\) $f(x) = 4 \cdot 3^x$

Explanation:

Step1: Substitute \(x = 1\) into \(f(x)=A\cdot b^{x}\)

We know that \(f(1) = 12\), so substituting \(x = 1\) into the function \(f(x)=A\cdot b^{x}\), we get \(f(1)=A\cdot b^{1}=A\cdot b = 12\). So we have the equation \(A\cdot b=12\), which can be rewritten as \(A=\frac{12}{b}\) (assuming \(b
eq0\)).

Step2: Substitute \(x = 3\) into \(f(x)=A\cdot b^{x}\)

We know that \(f(3)=108\), so substituting \(x = 3\) into the function \(f(x)=A\cdot b^{x}\), we get \(f(3)=A\cdot b^{3}=108\).

Step3: Substitute \(A=\frac{12}{b}\) into \(A\cdot b^{3}=108\)

Substitute \(A=\frac{12}{b}\) into \(A\cdot b^{3}=108\), we have \(\frac{12}{b}\cdot b^{3}=108\). Simplify the left - hand side: \(\frac{12}{b}\cdot b^{3}=12\cdot b^{2}\). So the equation becomes \(12b^{2}=108\).

Step4: Solve for \(b\)

Divide both sides of the equation \(12b^{2}=108\) by 12: \(b^{2}=\frac{108}{12} = 9\). Then take the square root of both sides. Since \(b>0\) (because it is the base of an exponential function), we get \(b = 3\).

Step5: Solve for \(A\)

Substitute \(b = 3\) into the equation \(A\cdot b=12\) (from Step 1). We have \(A\times3 = 12\), so \(A=\frac{12}{3}=4\).

Step6: Write the function

Now that we have \(A = 4\) and \(b = 3\), the exponential function is \(f(x)=4\cdot3^{x}\). We can also check the other options:

• For \(f(x)=\frac{4^{x}}{12}\), when \(x = 1\), \(f(1)=\frac{4}{12}=\frac{1}{3}

eq12\), so this option is wrong.

• For \(f(x)=12\cdot3^{x}\), when \(x = 1\), \(f(1)=12\times3 = 36

eq12\), so this option is wrong.

• For \(f(x)=49x^{- 37}\), this is a power function, not an exponential function of the form \(A\cdot b^{x}\), so this option is wrong.

Answer:

\(f(x)=4\cdot3^{x}\) (the option with the formula \(f(x) = 4\cdot3^{x}\))