Find the value of the constant k that makes the function continuous

Find the value of the constant k that makes the function continuous. 3x^2 – 8x – 16 if x < 4 kx – 4 if x = 4 g(x) Write an equation that can be solved to find k: 3x^2 – 8x – 16 = k(4) – 4 if x > 4 0 A. 842_84) – 18 = k(4) – 4 if x = 4 4 – 4 3(4)^2 – 8(4) – 16 = k(4) – 4 if x < 4 0 D. lim = k if x > 4 x -> 4 k =

The Correct Answer and Explanation is:

To find the value of the constant kkk that makes the function continuous at x=4x = 4x=4, we need to ensure that: lim⁡x→4−g(x)=g(4)=lim⁡x→4+g(x)\lim_{x \to 4^-} g(x) = g(4) = \lim_{x \to 4^+} g(x)x→4−lim​g(x)=g(4)=x→4+lim​g(x)

The piecewise function is defined as:

• g(x)=3×2−8x−16g(x) = 3x^2 – 8x – 16g(x)=3×2−8x−16 if x<4x < 4x<4
• g(x)=kx−4g(x) = kx – 4g(x)=kx−4 if x=4x = 4x=4
• g(x)=0g(x) = 0g(x)=0 if x>4x > 4x>4

Step 1: Evaluate the limit as x→4−x \to 4^-x→4−

Use the expression for x<4x < 4x<4: lim⁡x→4−g(x)=3(4)2−8(4)−16=48−32−16=0\lim_{x \to 4^-} g(x) = 3(4)^2 – 8(4) – 16 = 48 – 32 – 16 = 0x→4−lim​g(x)=3(4)2−8(4)−16=48−32−16=0

Step 2: Evaluate the limit as x→4+x \to 4^+x→4+

Use the expression for x>4x > 4x>4, which is a constant: lim⁡x→4+g(x)=0\lim_{x \to 4^+} g(x) = 0x→4+lim​g(x)=0

Step 3: Since both one-sided limits equal 0:

lim⁡x→4g(x)=0\lim_{x \to 4} g(x) = 0x→4lim​g(x)=0

To make the function continuous, the actual value at x=4x = 4x=4, that is g(4)g(4)g(4), must also equal 0. But from the definition: g(4)=k(4)−4=4k−4g(4) = k(4) – 4 = 4k – 4g(4)=k(4)−4=4k−4

Set this equal to the limit value: 4k−4=04k – 4 = 04k−4=0

Solve the equation:

4k=4k=14k = 4 \\ k = 14k=4k=1

Final Answer:

k=1k = 1k=1

Explanation

Continuity of a function at a specific point requires that the left-hand limit, the right-hand limit, and the function value at that point are all equal. In this problem, the function g(x)g(x)g(x) is piecewise-defined with different rules for when xxx is less than, equal to, or greater than 4.

For the function to be continuous at x=4x = 4x=4, we must ensure:

• lim⁡x→4−g(x)=lim⁡x→4+g(x)\lim_{x \to 4^-} g(x) = \lim_{x \to 4^+} g(x)limx→4−​g(x)=limx→4+​g(x)
• g(4)g(4)g(4) is equal to that common limit value

From the left, g(x)=3×2−8x−16g(x) = 3x^2 – 8x – 16g(x)=3×2−8x−16. Plugging in x=4x = 4x=4, we get: 3(4)2−8(4)−16=48−32−16=03(4)^2 – 8(4) – 16 = 48 – 32 – 16 = 03(4)2−8(4)−16=48−32−16=0

From the right, g(x)=0g(x) = 0g(x)=0 for x>4x > 4x>4, so: lim⁡x→4+g(x)=0\lim_{x \to 4^+} g(x) = 0x→4+lim​g(x)=0

Both limits equal 0, which means the overall limit exists and equals 0. Now, to make the function continuous, the function value at x=4x = 4x=4 must also be 0.

From the definition, g(4)=k(4)−4=4k−4g(4) = k(4) – 4 = 4k – 4g(4)=k(4)−4=4k−4. Set this equal to the limit: 4k−4=0⇒4k=4⇒k=14k – 4 = 0 \Rightarrow 4k = 4 \Rightarrow k = 14k−4=0⇒4k=4⇒k=1

Therefore, k=1k = 1k=1 ensures the function is continuous at x=4x = 4x=4.