Let the function $f : R o R$ is defined by $f(x) = x^2 + 3$, then $f^{-1}(12)$ is
(A) $\sqrt{3}$
(B) 3
(C) 9
(D) 12

Question

Let the function $f : R 	o R$ is defined by $f(x) = x^2 + 3$, then $f^{-1}(12)$ is
(A) $\sqrt{3}$
(B) 3
(C) 9
(D) 12

Accepted Answer

> [!TLDR]
> Solve $f(x)=12$: $x^2+3=12 \Rightarrow x^2=9 \Rightarrow x=3$.

### Step 1 — Meaning of $f^{-1}(12)$
$f^{-1}(12)$ is the input $x$ that the function maps to the output $12$, i.e. solve:

$$f(x) = 12 \;\Rightarrow\; x^2 + 3 = 12$$

### Step 2 — Solve for $x$

$$x^2 = 12 - 3 = 9 \;\Rightarrow\; x = \pm 3$$

### Step 3 — Select the value
Taking the principal (positive) root, $f^{-1}(12) = 3$, which is option (B).

**Why the other options are wrong:** (A) $\sqrt3$ solves $x^2=3$ (mis-set equation); (C) 9 is $x^2$, not $x$; (D) 12 is the output itself, not the pre-image. …

Step 1 — Meaning of f−1(12)f^{-1}(12)f−1(12)

Step 2 — Solve for xxx

Step 3 — Select the value

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Step 1 — Meaning of $f^{-1}(12)$

Step 2 — Solve for $x$

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