Question

Suppose g of x is equals to 1+ root x and f of g of x is equals to 3+2rootx+x then f of x

Answer 1

✪ᴀɴsᴡᴇʀ✪

• $\large{\boxed{\boxed{\green{ f(x) = x^2 + 2}}}}$

☆ɢɪᴠᴇɴ☆

• $g(x) = 1 + \sqrt{x}$

• $f(g(x)) = 3 + 2\sqrt{x} + x$

✵ᴛᴏ ғɪɴᴅ✵

• Expression of $f(x)$

✫ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ✫

Here,

$\to f(g(x)) = 3 + 2\sqrt{x} + x$

$\to f(g(x)) = 3 + 2\sqrt{x} + (\sqrt{x})^2$

$\to f(g(x)) = (\sqrt{x})^2 + 2\sqrt{x} + 1 + 2$

$\to f(g(x)) = (\sqrt{x}+ 1)^2 + 2$

$\to f(g(x)) = [g(x)]^2 + 2$

Put $g(x) = y$

$\to f(y) = y^2 + 2$

$\to f(x) = x^2 + 2$