1. Prove that the function f(x)=2x2−3x+2f(x) = 2x^2 - 3x + 2 is continuous at x=1x = 1. 2. The function f(x)f(x) is defined as follows : f(x)={2x−3,x<2x−1,x≥2f(x) = \begin{cases} 2x - 3, & x < 2 \\ x - 1, & x \ge 2 \end{cases} Prove that f(x)f(x) is continuous at x=2x = 2. 3. Discuss the continuity of the function f(x)={x+2,x≤1x−2,x>1f(x) = \begin{cases} x + 2, & x \le 1 \\ x - 2, & x > 1 \end{cases} at x=1x = 1.

Answer: For Question 2: Since lim⁡x→2−f(x)=lim⁡x→2+f(x)=f(2)=1\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 1, f(x)f(x) is continuous at x=2x = 2. For Question 3: Since lim⁡x→1−f(x)=3≠−1=lim⁡x→1+f(x)\lim_{x \to 1^-} f(x) = 3 \ne -1 = \lim_{x \to 1^+} f(x), f(x)f(x) is discontinuous at x=1x = 1.

Step-by-step solution

Step 1: Evaluate function value and limits for Question 2 at x=2x = 2

For Question 2, we evaluate the value of the function at x=2x = 2 using the branch x−1x - 1, giving f(2)=1f(2) = 1. Next, we evaluate the left-hand limit using 2x−32x - 3, which approaches 11, and the right-hand limit using x−1x - 1, which also approaches 11.

Step 2: Conclude continuity for Question 2

Since the left-hand limit, the right-hand limit, and the function value at x=2x = 2 are all equal to 11, the function f(x)f(x) is continuous at x=2x = 2. This completes the proof for Question 2.

Step 3: Calculate left-hand limit and function value for Question 3 at x=1x = 1

For Question 3, the function is defined as x+2x + 2 for x≤1x \le 1. Therefore, the value of the function at x=1x = 1 is 1+2=31 + 2 = 3, and the left-hand limit as xx approaches 11 from the left is also 33.

Step 4: Calculate right-hand limit for Question 3 at x=1x = 1

For values of x>1x > 1, the function rule is f(x)=x−2f(x) = x - 2. Taking the limit as xx approaches 11 from the right gives 1−2=−11 - 2 = -1.

Step 5: Determine continuity for Question 3

The left-hand limit is 33, while the right-hand limit is −1-1. Because the left-hand limit does not equal the right-hand limit, the limit of f(x)f(x) as x→1x \to 1 does not exist, so f(x)f(x) is discontinuous at x=1x = 1.

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