Question

consider the piecewise functions shown below.
f(x)=\begin{cases}\frac{1}{2}sqrt{x}+8, & 0leq xleq16\\-\frac{3}{2}|x - 22|+19, & 16lt xleq30end{cases}
g(x)=\begin{cases}-4sqrt{x}+18, & 0leq xleq9\\-2|x - 13|+17, & 9leq xleq18end{cases}
h(x)=\begin{cases}\frac{2}{3}|x - 12|+4, & 0leq xleq21\\5sqrt{x - 21}+11, & 21lt xleq30end{cases}
which of the above piecewise functions are continuous?

Explanation:

Step1: Check continuity for \(f(x)\) at \(x = 16\)

• Calculate \(\lim_{x

ightarrow16^{-}}f(x)\):

• Substitute \(x = 16\) into \(y=\frac{1}{2}\sqrt{x}+8\).
• \(y=\frac{1}{2}\sqrt{16}+8=\frac{1}{2}\times4 + 8=2 + 8=10\).
• Calculate \(\lim_{x

ightarrow16^{+}}f(x)\):

• Substitute \(x = 16\) into \(y =-\frac{3}{2}|x - 22|+19\).
• \(y=-\frac{3}{2}|16 - 22|+19=-\frac{3}{2}\times6+19=-9 + 19=10\).
• \(f(16)=\frac{1}{2}\sqrt{16}+8 = 10\). So \(f(x)\) is continuous at \(x = 16\).

Step2: Check continuity for \(g(x)\) at \(x = 9\)

• Calculate \(\lim_{x

ightarrow9^{-}}g(x)\):

• Substitute \(x = 9\) into \(y=-4\sqrt{x}+18\).
• \(y=-4\sqrt{9}+18=-4\times3+18=-12 + 18=6\).
• Calculate \(\lim_{x

ightarrow9^{+}}g(x)\):

• Substitute \(x = 9\) into \(y=-2|x - 13|+17\).
• \(y=-2|9 - 13|+17=-2\times4+17=-8 + 17=9\).
• Since \(\lim_{x

ightarrow9^{-}}g(x)
eq\lim_{x
ightarrow9^{+}}g(x)\), \(g(x)\) is not continuous at \(x = 9\).

Step3: Check continuity for \(h(x)\) at \(x = 21\)

• Calculate \(\lim_{x

ightarrow21^{-}}h(x)\):

• Substitute \(x = 21\) into \(y=\frac{2}{3}|x - 12|+4\).
• \(y=\frac{2}{3}|21 - 12|+4=\frac{2}{3}\times9+4=6 + 4=10\).
• Calculate \(\lim_{x

ightarrow21^{+}}h(x)\):

• Substitute \(x = 21\) into \(y=5\sqrt{x - 21}+11\).
• \(y=5\sqrt{21 - 21}+11=11\).
• Since \(\lim_{x

ightarrow21^{-}}h(x)
eq\lim_{x
ightarrow21^{+}}h(x)\), \(h(x)\) is not continuous at \(x = 21\).

Answer:

\(f(x)\) is continuous.