Q. Given k(x)={2x2−4 for x<2−3x+10 for x≥2a. Is k(x) continuous at x=2 ? Explain.b. If j(x)=k(x+1), what is the function for j(x) ?
Calculate k(2): question_prompt: Is k(x) continuous at x=2 and what is the function for j(x) if j(x)=k(x+1)?
Calculate left-hand limit: Calculate k(2) using the piece for x≥2: k(2)=−3(2)+10=−6+10=4.
Calculate right-hand limit: Calculate the limit of k(x) as x approaches 2 from the left using the piece for x < 2: limx→2−k(x)=2(22)−4=2(4)−4=8−4=4.
Check continuity at x=2: Since k(2)=4 and limx→2−k(x)=4, check the right-hand limit: limx→2+k(x)=−3(2)+10=−6+10=4.
Find j(x) for x<1: Since limx→2−k(x)=limx→2+k(x)=k(2)=4, k(x) is continuous at x=2.
Find j(x) for x≥1: To find j(x)=k(x+1), substitute x+1 for x in k(x). For x+1 < 2, x < 1: j(x)=2(x+1)2−4=2(x2+2x+1)−4=2x2+4x+2−4=2x2+4x−2.
Find j(x) for x≥1: To find j(x)=k(x+1), substitute x+1 for x in k(x). For x+1 < 2, x < 1: j(x)=2(x+1)2−4=2(x2+2x+1)−4=2x2+4x+2−4=2x2+4x−2. For x+1≥2, x≥10: x≥11.