To evaluate the function \( f(x) = x^2 - x + 2 \) for the given values, we can substitute each value of \( x \) into the function.
(a) \( f(3) \)
\[ f(3) = 3^2 - 3 + 2 = 9 - 3 + 2 = 8 \]
(b) \( f(0) \)
\[ f(0) = 0^2 - 0 + 2 = 0 - 0 + 2 = 2 \]
(c) \( f(-4) \)
\[ f(-4) = (-4)^2 - (-4) + 2 = 16 + 4 + 2 = 22 \]
(d) \( f\left(\frac{1}{4}\right) \)
\[ f\left(\frac{1}{4}\right) = \left(\frac{1}{4}\right)^2 - \frac{1}{4} + 2 = \frac{1}{16} - \frac{1}{4} + 2 \] To simplify \( -\frac{1}{4} \), we convert it to a fraction with a denominator of 16: \[ -\frac{1}{4} = -\frac{4}{16} \] So, \[ f\left(\frac{1}{4}\right) = \frac{1}{16} - \frac{4}{16} + 2 = \frac{1 - 4}{16} + 2 = -\frac{3}{16} + 2 \] Converting \( 2 \) to a fraction: \[ 2 = \frac{32}{16} \] Thus, \[ f\left(\frac{1}{4}\right) = -\frac{3}{16} + \frac{32}{16} = \frac{29}{16} \]
(e) \( f\left(-\frac{1}{7}\right) \)
\[ f\left(-\frac{1}{7}\right) = \left(-\frac{1}{7}\right)^2 - \left(-\frac{1}{7}\right) + 2 = \frac{1}{49} + \frac{1}{7} + 2 \] To simplify \( \frac{1}{7} \), we convert it to a fraction with a denominator of 49: \[ \frac{1}{7} = \frac{7}{49} \] So, \[ f\left(-\frac{1}{7}\right) = \frac{1}{49} + \frac{7}{49} + 2 = \frac{1 + 7}{49} + 2 = \frac{8}{49} + 2 \] Converting \( 2 \) to a fraction: \[ 2 = \frac{98}{49} \] Thus, \[ f\left(-\frac{1}{7}\right) = \frac{8}{49} + \frac{98}{49} = \frac{106}{49} \]
Summary of Results:
- \( f(3) = 8 \)
- \( f(0) = 2 \)
- \( f(-4) = 22 \)
- \( f\left(\frac{1}{4}\right) = \frac{29}{16} \)
- \( f\left(-\frac{1}{7}\right) = \frac{106}{49} \)