Porisma

Worked solution

Three ranges, three endpoint decisions

Find the range of each of the following functions.

(a) f:x↦x2+4x+3f:x\mapsto x^2+4x+3, x∈Rx\in\mathbb R, x≤−2x\leq-2.

(b) g:x↦e2xg:x\mapsto e^{2x}, x∈Rx\in\mathbb R, x>0x>0.

(c) h:x↦sin⁡xh:x\mapsto\sin x, x∈Rx\in\mathbb R, −π≤x≤π-\pi\leq x\leq\pi.

CJC · Functions · Self-Practice 3 · 2025

Compare your reasoning

(a) Complete the square: f(x)=(x+2)2−1f(x)=(x+2)^2-1. The minimum −1-1 occurs at the included input x=−2x=-2, and the function is unbounded above as x→−∞x\to-\infty. Therefore Rf=[−1,∞)R_f=[-1,\infty).

(b) Since x>0x>0, e2x>1e^{2x}>1. Every value above 1 is reached, but 1 is only approached as x→0+x\to0^+. Therefore Rg=(1,∞)R_g=(1,\infty).

(c) Both −1-1 and 1 are reached, at x=−π/2x=-\pi/2 and x=π/2x=\pi/2. Continuity supplies the values between them. Therefore Rh=[−1,1]R_h=[-1,1].

How did this question go?

Common mistakes

Checking only the two ends of the input interval can miss an interior maximum or minimum. An excluded input endpoint does not automatically exclude its output if another allowed input reaches that value.