Given the function [tex]\( f(x) = -2x^3 + 36x^2 - 162x + 1 \)[/tex], we need to find its local minimum and maximum points.
### Step-by-Step Solution
1. Finding the First Derivative:
To determine the critical points of the function, we need to find the first derivative [tex]\( f'(x) \)[/tex].
[tex]\( f'(x) = \frac{d}{dx} (-2x^3 + 36x^2 - 162x + 1) \)[/tex]
Using the power rule for differentiation, we get:
[tex]\( f'(x) = -6x^2 + 72x - 162 \)[/tex]
2. Setting the First Derivative to Zero:
To find the critical points, we solve the equation [tex]\( f'(x) = 0 \)[/tex].
[tex]\[ -6x^2 + 72x - 162 = 0 \][/tex]
Dividing the entire equation by [tex]\(-6\)[/tex]:
[tex]\[ x^2 - 12x + 27 = 0 \][/tex]
This is a quadratic equation that can be solved using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = 27 \)[/tex].
3. Solving the Quadratic Equation:
[tex]\[ x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 27}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{144 - 108}}{2} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ x = \frac{12 \pm 6}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{12 + 6}{2} = \frac{18}{2} = 9 \][/tex]
[tex]\[ x = \frac{12 - 6}{2} = \frac{6}{2} = 3 \][/tex]
So, the critical points are [tex]\( x = 9 \)[/tex] and [tex]\( x = 3 \)[/tex].
4. Finding the Second Derivative:
We now need to use the second derivative test to classify these critical points as minima or maxima.
[tex]\( f''(x) = \frac{d}{dx} (-6x^2 + 72x - 162) \)[/tex]
Using the power rule again:
[tex]\[ f''(x) = -12x + 72 \][/tex]
5. Evaluating the Second Derivative at Critical Points:
We substitute the critical points into the second derivative:
- For [tex]\( x = 9 \)[/tex]:
[tex]\[ f''(9) = -12(9) + 72 = -108 + 72 = -36 \][/tex]
Since [tex]\( f''(9) < 0 \)[/tex], the function has a local maximum at [tex]\( x = 9 \)[/tex].
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ f''(3) = -12(3) + 72 = -36 + 72 = 36 \][/tex]
Since [tex]\( f''(3) > 0 \)[/tex], the function has a local minimum at [tex]\( x = 3 \)[/tex].
6. Finding the Function Values at Critical Points:
We substitute the critical points back into the original function to find the corresponding function values:
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = -2(3)^3 + 36(3)^2 - 162(3) + 1 \][/tex]
[tex]\[ f(3) = -2(27) + 36(9) - 162(3) + 1 \][/tex]
[tex]\[ f(3) = -54 + 324 - 486 + 1 \][/tex]
[tex]\[ f(3) = -215 \][/tex]
- For [tex]\( x = 9 \)[/tex]:
[tex]\[ f(9) = -2(9)^3 + 36(9)^2 - 162(9) + 1 \][/tex]
[tex]\[ f(9) = -2(729) + 36(81) - 162(9) + 1 \][/tex]
[tex]\[ f(9) = -1458 + 2916 - 1458 + 1 \][/tex]
[tex]\[ f(9) = 1 \][/tex]
So, the function has:
- A local minimum at [tex]\( x = 3 \)[/tex] with value [tex]\( -215 \)[/tex].
- A local maximum at [tex]\( x = 9 \)[/tex] with value [tex]\( 1 \)[/tex].
