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smc

algebra senior

Problem

If

f (x) = 3 x + 2

for all real

x

, then the statement: "

∣ f (x) + 4∣ < a

whenever

∣ x + 2∣ < b

and

a > 0

and

b > 0

" is true when

(A)

b l t . e q a \/ 3

(B)

b > a \/ 3

(C)

a l t . e q b \/ 3

(D)

a > b \/ 3

Solution

Plugging in

f (x) = 3 x + 2

into

∣ f (x) + 4∣ < a

, we get

∣3 x + 6∣ = ∣3 (x + 2) ∣ = 3∣ x + 2∣ < a

. Therefore,

∣ x + 2∣ < \frac{a}{3}

. Therefore, we want

∣ x + 2∣ < b ⟹ ∣ x + 2∣ < \frac{a}{3}

. This is clearly only true when

b \leq \frac{a}{3}, b \leq a /3

.

Final answer

A