Porisma

Worked solution

Apply the vertical-line test

State and justify which of the following is a function. (a) f:x↦x2+1f:x\mapsto x^2+1, x∈Rx\in\mathbb R. (b) g:x↦±xg:x\mapsto\pm\sqrt{x}, x∈Rx\in\mathbb R, x≥0x\ge0.

CJC · Functions · Example 2

(a) One ordinate for every input

Answer

f is a function.

The source parabola has vertex (0,1)(0,1). Every vertical line meets its graph at most once, so each input has exactly one output.

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Upward parabola y=x squared+1, vertex (0,1). Each vertical line meets it once.
Upward parabola y=x squared+1, vertex (0,1). Each vertical line meets it once.
Upward parabola y=x squared+1, vertex (0,1). Each vertical line meets it once.
How did this part go?

(b) A single counterexample is enough

Answer

g is not a function.

The relation has both branches y=xy=\sqrt x and y=−xy=-\sqrt x. The Teacher adds the vertical line x=2x=2, which meets both branches. Independently those outputs are 2\sqrt2 and −2-\sqrt2, two distinct values for one allowed input. At zero the branches meet; that does not repair the failure elsewhere.

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Both branches y=±square root x for x nonnegative. Teacher line x=2 meets them at ±square root 2.
Both branches y=±square root x for x nonnegative. Teacher line x=2 meets them at ±square root 2.
Both branches y=±square root x for x nonnegative. Teacher line x=2 meets them at ±square root 2.
How did this part go?

Solution notes

The radical is present in both original page images but lost in the native text extraction. The original source uses g as a candidate relation to be tested, not as an assertion that it is already a function.