Find fog and gof when f(x) = 2x+1 and g(x) = x²

Question

If $f(x) = 2x + 1$ and $g(x) = x^2$ , find: (a) $fog(x)$ — also written as $f \circ g(x)$ or $f(g(x))$ (b) $gof(x)$ — also written as $g \circ f(x)$ or $g(f(x))$

Also verify whether $fog = gof$ .

Solution — Step by Step

$fog(x)$ means: first apply $g$ , then apply $f$ to the result. The notation reads right-to-left: in $f \circ g$ , $g$ acts first.

$gof(x)$ means: first apply $f$ , then apply $g$ to the result. In $g \circ f$ , $f$ acts first.

A useful way to remember: $fog(x) = f(g(x))$ — substitute $g(x)$ wherever you see $x$ in $f$ .

Start with $g(x) = x^2$ . Now substitute this into $f$ :

fog(x) = f(g(x)) = f(x^2)

Replace $x$ with $x^2$ in $f(x) = 2x + 1$ :

fog(x) = 2(x^2) + 1 = \mathbf{2x^2 + 1}

Start with $f(x) = 2x + 1$ . Now substitute this into $g$ :

gof(x) = g(f(x)) = g(2x + 1)

Replace $x$ with $(2x + 1)$ in $g(x) = x^2$ :

gof(x) = (2x + 1)^2 = 4x^2 + 4x + 1 = \mathbf{4x^2 + 4x + 1}

$fog(x) = 2x^2 + 1$ and $gof(x) = 4x^2 + 4x + 1$ .

These are clearly different expressions, so $fog \neq gof$ .

This confirms that function composition is generally not commutative.

Why This Works

Function composition is essentially substitution — you’re plugging one function’s output as the other function’s input. The order matters because the transformations are different: $fog$ first squares $x$ then applies the linear transformation; $gof$ first applies the linear transformation then squares.

Think of it as two machines in a factory. $fog$ : raw material → Machine G (squaring) → Machine F (double and add 1). $gof$ : raw material → Machine F first → Machine G. Different orders produce different products.

Alternative Method

Verify with a specific value. Let $x = 2$ :

$fog(2) = f(g(2)) = f(4) = 2(4) + 1 = 9$ . Check: $2(2)^2 + 1 = 9$ ✓

$gof(2) = g(f(2)) = g(5) = 25$ . Check: $4(4) + 4(2) + 1 = 16 + 8 + 1 = 25$ ✓

Substituting a specific value is a quick way to verify your algebra for composition problems.

For CBSE Class 12 and JEE, always write out the full substitution step clearly. In exams, the marks are awarded for showing $f(g(x)) = f(\text{expression for } g(x))$ and then simplifying — not just writing the final answer.

Common Mistake

The most frequent error is reversing the order: students compute $fog$ but actually perform $gof$ . Remember: $fog = f(g(x))$ — in this notation, ” $g$ is inside $f$ ,” meaning $g$ goes first. If you’re unsure, always expand: $fog(x) = f(g(x))$ , and now it’s clear — evaluate $g$ first, then put the result into $f$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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