Answer
$\frac{65}{24}$
Work Step by Step
$P_{4}(x) = 1 + 4x + \frac{16}{2!}x^{2} + \frac{64}{3!}x^{3} + \frac{256}{4!}x^{4}$
$f(\frac{1}{4})$, so substitute $\frac{1}{4}$ for $x$.
$P_{4}(\frac{1}{4}) = 1 + 4(\frac{1}{4}) + \frac{16}{2!}(\frac{1}{4})^{2} + \frac{64}{3!}(\frac{1}{4})^{3} + \frac{256}{4!}(\frac{1}{4})^{4}$
$= 1+ 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24}$
$=\frac{65}{24}$
