Christmas Streak 52/88: Calculus 1 Basics #2

Calculus Level 3

A continuous function f ( x ) f(x) satisfies the conditions below for a constant a . a.

{ f ( x ) = x 2 − 6 x + 8 for − 1 ≤ x < a − 1 f ( x ) = f ( x + a ) \begin{cases} f(x) =x^2-6x+8 & \text{for } -1\le x < a-1 \\\\ f(x)=f(x+a) \end{cases}

Find the value of the positive real a . a.

This problem is a part of <Christmas Streak 2017> series .

8 5 7 9 6

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Nov 22, 2017

lim ⁡ x → − 1 − f ( x ) = f ( − 1 ) lim ⁡ x → a − 1 − f ( x ) = 15 ( a − 1 ) 2 − 6 ( a − 1 ) + 8 = 15 ( a − 1 ) 2 − 6 ( a − 1 ) − 7 = 0 { ( a − 1 ) + 1 } { ( a − 1 ) − 7 } = 0 a = 0 or 8 \lim_{x\to -1^{-}}f(x)=f(-1) \\ \\ \lim_{x\to a-1^{-}} f(x)=15 \\ \\ (a-1)^2-6(a-1)+8=15 \\ \\ (a-1)^2-6(a-1)-7=0 \\ \\ \{(a-1)+1\}\{(a-1)-7\}=0 \\ \\ a=0~\text{or}~8

Since a a is positive, a = 8 . \boxed{a=8}.

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