To solve the problem, we need to find the local maximum and minimum values of the function [tex]\( y = x^5 - x^4 + x^3 + x^2 - x + 2 \)[/tex] and evaluate the function at these points.
### Step 1: Calculating the first derivative
The first step is to find the first derivative [tex]\( f'(x) \)[/tex] of the given function [tex]\( f(x) = x^5 - x^4 + x^3 + x^2 - x + 2 \)[/tex].
Using standard differentiation rules:
[tex]\[ f'(x) = \frac{d}{dx}(x^5 - x^4 + x^3 + x^2 - x + 2) \][/tex]
[tex]\[ f'(x) = 5x^4 - 4x^3 + 3x^2 + 2x - 1 \][/tex]
### Step 2: Finding critical points
Next, we find the critical points by setting the first derivative equal to zero and solving for [tex]\( x \)[/tex]:
[tex]\[ 5x^4 - 4x^3 + 3x^2 + 2x - 1 = 0 \][/tex]
### Step 3: Solving the derivative equation
Solving this polynomial equation can be complex, but for now, assume we have the critical points [tex]\( x = -0.558 \)[/tex] and [tex]\( x = 0.358 \)[/tex] based on provided information.
### Step 4: Evaluating the function at critical points
To find the function's value at these critical points:
#### Local Maximum at [tex]\( x = -0.558 \)[/tex]
[tex]\[ f(-0.558) = (-0.558)^5 - (-0.558)^4 + (-0.558)^3 + (-0.558)^2 - (-0.558) + 2 \][/tex]
Calculating each term:
[tex]\[
\begin{align*}
(-0.558)^5 & = -0.0556 \\
(-0.558)^4 & = 0.0973 \\
(-0.558)^3 & = -0.174 \\
(-0.558)^2 & = 0.311 \\
-(-0.558) & = 0.558 \\
\end{align*}
\][/tex]
Adding them together:
[tex]\[ f(-0.558) = -0.0556 - 0.0973 - 0.174 + 0.311 + 0.558 + 2 \approx 2.5421 \][/tex]
Rounded to three decimal places:
[tex]\[ f(-0.558) \approx 2.542 \][/tex]
#### Local Minimum at [tex]\( x = 0.358 \)[/tex]
[tex]\[ f(0.358) = (0.358)^5 - (0.358)^4 + (0.358)^3 + (0.358)^2 - (0.358) + 2 \][/tex]
Calculating each term:
[tex]\[
\begin{align*}
(0.358)^5 & = 0.0042 \\
(0.358)^4 & = 0.0165 \\
(0.358)^3 & = 0.046 \\
(0.358)^2 & = 0.128 \\
-(0.358) & = -0.358 \\
\end{align*}
\][/tex]
Adding them together:
[tex]\[ f(0.358) = 0.0042 - 0.0165 + 0.046 + 0.128 - 0.358 + 2 \approx 1.8037 \][/tex]
Rounded to three decimal places:
[tex]\[ f(0.358) \approx 1.804 \][/tex]
### Conclusion
The function has a local maximum at the point [tex]\((-0.558, 2.542)\)[/tex] and a local minimum at the point [tex]\((0.358, 1.804)\)[/tex].
So the filled parts:
- Local maximum at the point [tex]\((-0.558, 2.542)\)[/tex]
- Local minimum at the point [tex]\((0.358, 1.804)\)[/tex]
