Question 7 Let the function f(x)f(x) be defined by f(x)={x2−4,x<2kx+1,x≥2f(x) = \begin{cases} x^2 - 4, & x < 2 \\ kx + 1, & x \geq 2 \end{cases} Determine the value of kk that makes f(x)f(x) continuous at x=2x = 2. Show solutionHide solution+Question 7 - Solution We are given a piecewise function: f(x)={x2−4,x<2kx+1,x≥2f(x) = \begin{cases} x^2 - 4, & x < 2 \\ kx + 1, & x \geq 2 \end{cases} To ensure continuity at x=2x = 2, we require: limx→2−f(x)=limx→2+f(x)=f(2)\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) Evaluate the left-hand limit: limx→2−f(x)=limx→2−(x2−4)=4−4=0\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2 - 4) = 4 - 4 = 0 Evaluate the right-hand limit: limx→2+f(x)=limx→2+(kx+1)=2k+1\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (kx + 1) = 2k + 1 Set the two limits equal for continuity: 2k+1=0⇒k=−122k + 1 = 0 \quad \Rightarrow \quad k = -\frac{1}{2} Final Answer: k=−12\boxed{k = -\dfrac{1}{2}}
