Question

a coordinate on the graph of $f(x) = \sqrt3{x}$ is $(64, 4)$. the function is transformed into $f(x) = k\sqrt3{x}$, and a point on the new graph is $(64, 2)$. what is the value of $k$? (1 point)
\bigcirc $\frac{1}{2}$
\bigcirc $4$
\bigcirc $64$
\bigcirc $2$

Explanation:

Step1: Substitute the point into the new function

We know that for the new function \( f(x) = k\sqrt[3]{x} \), and the point \((64, 2)\) lies on its graph. So we substitute \( x = 64 \) and \( f(x)=2 \) into the function.
We get the equation \( 2 = k\sqrt[3]{64} \).

Step2: Calculate the cube root of 64

We know that \( \sqrt[3]{64} = 4 \) because \( 4\times4\times4 = 64 \). So the equation becomes \( 2 = k\times4 \).

Step3: Solve for k

To solve for \( k \), we divide both sides of the equation \( 2 = 4k \) by 4.
\( k=\frac{2}{4}=\frac{1}{2} \).

Answer:

\(\frac{1}{2}\) (corresponding to the option "A. \(\frac{1}{2}\)" assuming the first option is labeled as such)