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Given 
k(x)={[2x^(2)-4" for "x < 2],[-3x+10" for "x >= 2]:}
a. Is 
k(x) continuous at 
x=2 ? Explain.
b. If 
j(x)=k(x+1), what is the function for 
j(x) ?

Given k(x)={2x24 for xlt;23x+10 for x2 k(x)=\left\{\begin{array}{c}2 x^{2}-4 \text { for } x&lt;2 \\ -3 x+10 \text { for } x \geq 2\end{array}\right. \newlinea. Is k(x) k(x) continuous at x=2 x=2 ? Explain.\newlineb. If j(x)=k(x+1) j(x)=k(x+1) , what is the function for j(x) j(x) ?

Full solution

Q. Given k(x)={2x24 for x<23x+10 for x2 k(x)=\left\{\begin{array}{c}2 x^{2}-4 \text { for } x<2 \\ -3 x+10 \text { for } x \geq 2\end{array}\right. \newlinea. Is k(x) k(x) continuous at x=2 x=2 ? Explain.\newlineb. If j(x)=k(x+1) j(x)=k(x+1) , what is the function for j(x) j(x) ?
  1. Calculate k(2)k(2): question_prompt: Is k(x)k(x) continuous at x=2x=2 and what is the function for j(x)j(x) if j(x)=k(x+1)j(x)=k(x+1)?
  2. Calculate left-hand limit: Calculate k(2)k(2) using the piece for x2x \geq 2: k(2)=3(2)+10=6+10=4k(2) = -3(2) + 10 = -6 + 10 = 4.
  3. Calculate right-hand limit: Calculate the limit of k(x)k(x) as xx approaches 22 from the left using the piece for x < 2: limx2k(x)=2(22)4=2(4)4=84=4\lim_{x\to2^-} k(x) = 2(2^2) - 4 = 2(4) - 4 = 8 - 4 = 4.
  4. Check continuity at x=2x=2: Since k(2)=4k(2) = 4 and limx2k(x)=4\lim_{x\to2-} k(x) = 4, check the right-hand limit: limx2+k(x)=3(2)+10=6+10=4\lim_{x\to2+} k(x) = -3(2) + 10 = -6 + 10 = 4.
  5. Find j(x)j(x) for x<1: Since limx2k(x)=limx2+k(x)=k(2)=4\lim_{x\to 2^-} k(x) = \lim_{x\to 2^+} k(x) = k(2) = 4, k(x)k(x) is continuous at x=2x = 2.
  6. Find j(x)j(x) for x1x\geq1: To find j(x)=k(x+1)j(x) = k(x+1), substitute x+1x+1 for xx in k(x)k(x). For x+1 < 2, x < 1: j(x)=2(x+1)24=2(x2+2x+1)4=2x2+4x+24=2x2+4x2j(x) = 2(x+1)^2 - 4 = 2(x^2 + 2x + 1) - 4 = 2x^2 + 4x + 2 - 4 = 2x^2 + 4x - 2.
  7. Find j(x)j(x) for x1x\geq1: To find j(x)=k(x+1)j(x) = k(x+1), substitute x+1x+1 for xx in k(x)k(x). For x+1 < 2, x < 1: j(x)=2(x+1)24=2(x2+2x+1)4=2x2+4x+24=2x2+4x2j(x) = 2(x+1)^2 - 4 = 2(x^2 + 2x + 1) - 4 = 2x^2 + 4x + 2 - 4 = 2x^2 + 4x - 2. For x+12x+1 \geq 2, x1x\geq100: x1x\geq111.

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