To evaluate the function \( f(x) = 4x + 1 \) at the given values, let's solve each:
a. \( f(-9) \)
To find \( f(-9) \), substitute \(-9\) for \(x\):
\[ f(-9) = 4(-9) + 1 \] \[ = -36 + 1 \] \[ = -35 \]
So, \[ \text{a. } f(-9) = -35 \]
b. \( f(x+5) \)
To find \( f(x+5) \), substitute \(x + 5\) for \(x\):
\[ f(x+5) = 4(x+5) + 1 \] \[ = 4x + 20 + 1 \] \[ = 4x + 21 \]
So, \[ \text{b. } f(x+5) = 4x + 21 \]
c. \( f(-x) \)
To find \( f(-x) \), substitute \(-x\) for \(x\):
\[ f(-x) = 4(-x) + 1 \] \[ = -4x + 1 \]
So, \[ \text{c. } f(-x) = -4x + 1 \]
Summary of Results:
- a. \( f(-9) = -35 \)
- b. \( f(x+5) = 4x + 21 \)
- c. \( f(-x) = -4x + 1 \)