Evaluate The Piecewise Function At The Given Values Of The Independent Variable. If X #4 X² – 16 H(X) = {X-4 7 If X = 4 (A) H(3) (B) H(0) (C) H(4) (A) H(3)= (B) H(0) = (C) H(4)=1 The Domain Of The Piecewise Function Is ( -00,00) A. Graph The Function B. Use Your Graph To Determine The Function’s Range F(X) = Of X-4 – 2x – 45x
The Correct Answer and Explanation is:
Given Piecewise Function:
H(x)={x2−16x−4,if x≠47,if x=4H(x) = \begin{cases} \frac{x^2 – 16}{x – 4}, & \text{if } x \ne 4 \\ 7, & \text{if } x = 4 \end{cases}
We are asked to find:
- (A) H(3)H(3)
- (B) H(0)H(0)
- (C) H(4)H(4)
- Domain
- Graph the function
- Use the graph to find the range
✅ Step-by-Step Evaluations:
(A) H(3)H(3)
Since x=3≠4x = 3 \ne 4, use: H(x)=x2−16x−4H(x) = \frac{x^2 – 16}{x – 4} H(3)=32−163−4=9−16−1=−7−1=7H(3) = \frac{3^2 – 16}{3 – 4} = \frac{9 – 16}{-1} = \frac{-7}{-1} = 7
(B) H(0)H(0)
Since x=0≠4x = 0 \ne 4, use the same expression: H(0)=02−160−4=−16−4=4H(0) = \frac{0^2 – 16}{0 – 4} = \frac{-16}{-4} = 4
(C) H(4)H(4)
Since x=4x = 4, we use the second part of the function: H(4)=7H(4) = 7
✅ Final Answers:
- (A) H(3)=7H(3) = 7
- (B) H(0)=4H(0) = 4
- (C) H(4)=7H(4) = 7
🌐 Domain of H(x)H(x):
The function is defined for all real numbers, including x=4x = 4, because it assigns a separate value at that point.
So the domain is:
(−∞,∞)\boxed{(-\infty, \infty)}
📉 Graph and Explanation:
To graph the function, consider the simplified form of the first part. x2−16x−4=(x−4)(x+4)x−4=x+4,for x≠4\frac{x^2 – 16}{x – 4} = \frac{(x – 4)(x + 4)}{x – 4} = x + 4, \quad \text{for } x \ne 4
So the graph is:
- A line y=x+4y = x + 4, with a hole at x=4x = 4
- At x=4x = 4, the function jumps to y=7y = 7
This is a piecewise discontinuous function:
- A straight line with a hole at (4,8)(4, 8)
- A filled dot at (4,7)(4, 7)
📊 Range of H(x)H(x):
The line y=x+4y = x + 4 covers all real numbers except at x=4x = 4, where it jumps from y=8y = 8 to y=7y = 7.
So the output at x=4x = 4 is 7, but y=8y = 8 is never taken because of the hole.
The range is:
(−∞,8)∪{7}∪(8,∞)\boxed{(-\infty, 8) \cup \{7\} \cup (8, \infty)}
We include 7 because H(4)=7H(4) = 7, but exclude 8 because the function never equals 8.
