Question

the function f is defined as follows.
$f(x)=\sqrt3{x} + 1$
find $f(64)$ and $f(-8)$.
$f(64) = \square$
$f(-8) = \square$

Explanation:

Step1: Find \( f(64) \)

Substitute \( x = 64 \) into \( f(x)=\sqrt[3]{x}+1 \). We know that \( \sqrt[3]{64} = 4 \) because \( 4^3 = 64 \). Then \( f(64)=\sqrt[3]{64}+1 = 4 + 1 \).

Step2: Calculate \( f(64) \)

\( 4 + 1 = 5 \), so \( f(64)=5 \).

Step3: Find \( f(-8) \)

Substitute \( x = -8 \) into \( f(x)=\sqrt[3]{x}+1 \). We know that \( \sqrt[3]{-8} = -2 \) because \( (-2)^3 = -8 \). Then \( f(-8)=\sqrt[3]{-8}+1 = -2 + 1 \).

Step4: Calculate \( f(-8) \)

\( -2 + 1 = -1 \), so \( f(-8)=-1 \).

Answer:

\( f(64) = 5 \)
\( f(-8) = -1 \)