Question

1. given f(x) = {12 / (x + 3), x < 1; 5x - 2, 1 ≤ x < 7; |4x - 8|, x ≥ 7} a) f(7) b) the value(s) of x, at which f(x) is not continuous

Explanation:

Step1: Recall the definition of continuity

A function \(y = f(x)\) is continuous at \(x = a\) if \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)\). We need to check the continuity at the break - points of the piece - wise function. The break - points of \(f(x)\) are \(x = 1\) and \(x=7\).

Step2: Check continuity at \(x = 1\)

For \(x<1\), \(f(x)=\frac{12}{x + 3}\). For \(1\leq x<7\), \(f(x)=5x-2\).
\(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}\frac{12}{x + 3}=\frac{12}{1 + 3}=3\).
\(\lim_{x
ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}(5x - 2)=5\times1-2 = 3\). And \(f(1)=5\times1-2=3\). So \(f(x)\) is continuous at \(x = 1\).

Step3: Check continuity at \(x = 7\)

For \(1\leq x<7\), \(f(x)=5x-2\). For \(x\geq7\), \(f(x)=|4x-3|-8\).
\(\lim_{x
ightarrow7^{-}}f(x)=\lim_{x
ightarrow7^{-}}(5x - 2)=5\times7-2=33\).
For \(x\geq7\), \(f(x)=|4x - 3|-8\). When \(x\geq7\), \(4x-3\geq4\times7-3=25>0\), so \(f(x)=4x-3 - 8=4x-11\).
\(\lim_{x
ightarrow7^{+}}f(x)=\lim_{x
ightarrow7^{+}}(4x-11)=4\times7-11=17\). Since \(\lim_{x
ightarrow7^{-}}f(x)=33\) and \(\lim_{x
ightarrow7^{+}}f(x)=17\), \(f(x)\) is not continuous at \(x = 7\).

Step4: Calculate \(f(7)\)

Since for \(x\geq7\), \(f(x)=|4x - 3|-8\), when \(x = 7\), \(4x-3=4\times7-3 = 25>0\), so \(f(7)=4\times7-11=17\).

Answer:

a) \(17\)
b) \(x = 7\)