10. 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…

Explanation:

Step1: Recall the definition of continuity

A function (y = f(x)) is continuous at (x = a) if (\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow 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\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{-}}\frac{12}{x + 3}=\frac{12}{1 + 3}=3). (\lim_{x\rightarrow1^{+}}f(x)=\lim_{x\rightarrow1^{+}}(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\rightarrow7^{-}}f(x)=\lim_{x\rightarrow7^{-}}(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\rightarrow7^{+}}f(x)=\lim_{x\rightarrow7^{+}}(4x-11)=4\times7-11=17). Since (\lim_{x\rightarrow7^{-}}f(x)=33) and (\lim_{x\rightarrow7^{+}}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)