Question

(2) if : $f(x) = \sqrt3{x + 5}$, $g(x) = x^2$, then : $(f \circ g)(2) = \dots \dots \dots \dots \dots$
(a) 7
(b) 3
(c) 4
(d) 9

Explanation:

Step1: Find \( g(2) \)

First, we need to find the value of \( g(2) \) since \( (f \circ g)(2)=f(g(2)) \). Given \( g(x)=x^2 \), substitute \( x = 2 \) into \( g(x) \):
\( g(2)=2^2 = 4 \)

Step2: Find \( f(g(2)) \)

Now, we substitute \( g(2)=4 \) into \( f(x) \). Given \( f(x)=\sqrt[3]{x + 5} \), substitute \( x = 4 \) into \( f(x) \):
\( f(4)=\sqrt[3]{4 + 5}=\sqrt[3]{9} \)? Wait, no, wait. Wait, \( 4 + 5 = 9 \)? Wait, no, wait, \( g(2)=4 \), so \( f(g(2))=f(4)=\sqrt[3]{4 + 5}=\sqrt[3]{9} \)? But that's not one of the options. Wait, maybe I made a mistake. Wait, the options are 7, 3, 4, 9. Wait, maybe the function is \( f(x)=\sqrt{x + 5} \) (square root) instead of cube root? Let me check again. If \( f(x)=\sqrt{x + 5} \) and \( g(x)=x^2 \), then \( g(2)=4 \), so \( f(4)=\sqrt{4 + 5}=\sqrt{9}=3 \). Ah, that must be it. Maybe a typo in the cube root, should be square root. So assuming \( f(x)=\sqrt{x + 5} \) (since the options make sense with square root):

Step1: Find \( g(2) \)

\( g(2)=2^2 = 4 \)

Step2: Find \( f(g(2)) \)

\( f(4)=\sqrt{4 + 5}=\sqrt{9}=3 \)

Answer:

b. 3