Answer:
To find the relative maxima and minima, we need to find the critical points by taking the derivative of the function and setting it equal to zero.
Given function: \( f(x) = -2x^4 + x - 1 \)
First, let's find the derivative:
\[ f'(x) = -8x^3 + 1 \]
Now, set the derivative equal to zero and solve for \(x\):
\[ -8x^3 + 1 = 0 \]
\[ -8x^3 = -1 \]
\[ x^3 = \frac{1}{8} \]
\[ x = \sqrt[3]{\frac{1}{8}} \]
\[ x = \frac{1}{2} \]
Now, we need to determine the nature of the critical point:
\[ f''(x) = -24x^2 \]
Evaluate \( f''(x) \) at \( x = \frac{1}{2} \):
\[ f''\left(\frac{1}{2}\right) = -24\left(\frac{1}{2}\right)^2 \]
\[ f''\left(\frac{1}{2}\right) = -6 \]
Since \( f''\left(\frac{1}{2}\right) \) is negative, \( x = \frac{1}{2} \) is a relative maximum.
To find the absolute maximum, we need to evaluate the function at the critical points and at the endpoints of the domain.
\[ f\left(\frac{1}{2}\right) = -2\left(\frac{1}{2}\right)^4 + \frac{1}{2} - 1 \]
\[ f\left(\frac{1}{2}\right) = -\frac{1}{16} + \frac{1}{2} - 1 \]
\[ f\left(\frac{1}{2}\right) = -\frac{1}{16} + \frac{8}{16} - \frac{16}{16} \]
\[ f\left(\frac{1}{2}\right) = -\frac{9}{16} \]
Since \( f\left(\frac{1}{2}\right) = -\frac{9}{16} \) is the only critical point, it is also the absolute maximum.
For the zero/multiplicity:
The function \( f(x) = -2x^4 + x - 1 \) does not have any real zeros because it's a polynomial function with no linear or constant term.
For intervals of increase/decrease:
Since the function has a relative maximum at \( x = \frac{1}{2} \), it decreases for \( x < \frac{1}{2} \) and increases for \( x > \frac{1}{2} \).
