Question

find the values of ( k ) that make the function continuous. f(x) = \begin{cases} k^2 - 18x, & x > 3 \\ kx, & x leq 3 end{cases}

Explanation:

Step1: Recall the definition of continuity at a point

For a function \( f(x) \) to be continuous at \( x = a \), the following must hold: \( \lim_{x \to a^-} f(x)=\lim_{x \to a^+} f(x)=f(a) \). Here, we need continuity at \( x = 3 \). So we need to find \( \lim_{x \to 3^-} f(x) \), \( \lim_{x \to 3^+} f(x) \) and set them equal to \( f(3) \).

Step2: Calculate the left - hand limit (\( x\to3^- \))

When \( x\leq3 \), \( f(x)=kx \). So \( \lim_{x \to 3^-} f(x)=\lim_{x \to 3^-} kx \). Using the limit rule for a linear function \( \lim_{x\to a}mx = ma \) (where \( m = k \) and \( a = 3 \)), we get \( \lim_{x \to 3^-} kx=k\times3 = 3k \). Also, \( f(3)=k\times3 = 3k \) (since \( x = 3 \) satisfies \( x\leq3 \)).

Step3: Calculate the right - hand limit (\( x\to3^+ \))

When \( x > 3 \), \( f(x)=k^{2}-18x \). So \( \lim_{x \to 3^+} f(x)=\lim_{x \to 3^+}(k^{2}-18x) \). Using the limit rule \( \lim_{x\to a}(c - mx)=c - ma \) (where \( c = k^{2} \), \( m = 18 \) and \( a = 3 \)), we have \( \lim_{x \to 3^+}(k^{2}-18x)=k^{2}-18\times3=k^{2}-54 \).

Step4: Set the left - hand limit equal to the right - hand limit

For the function to be continuous at \( x = 3 \), \( \lim_{x \to 3^-}f(x)=\lim_{x \to 3^+}f(x) \). So we set \( 3k=k^{2}-54 \).

Step5: Solve the quadratic equation

Rearrange the equation \( 3k = k^{2}-54 \) to get a quadratic equation in standard form: \( k^{2}-3k - 54=0 \).
We factor the quadratic equation. We need two numbers that multiply to \( - 54 \) and add up to \( - 3 \). The numbers are \( - 9 \) and \( 6 \) since \( (-9)\times6=-54 \) and \( - 9 + 6=-3 \).
So, \( k^{2}-3k - 54=(k - 9)(k + 6)=0 \).
Setting each factor equal to zero gives \( k - 9 = 0 \) or \( k+6 = 0 \).
Solving for \( k \), we get \( k = 9 \) or \( k=-6 \).

Answer:

The values of \( k \) that make the function continuous are \( k = 9 \) and \( k=-6 \).