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Sagot :
To determine the local minimum and maximum of the function [tex]\( f(x) = -2x^3 + 30x^2 - 54x + 4 \)[/tex], we follow these steps:
### Step 1: Find the First Derivative
The first derivative of the function [tex]\( f(x) \)[/tex], denoted as [tex]\( f'(x) \)[/tex], is found using the power rule:
[tex]\[ f'(x) = \frac{d}{dx} (-2x^3 + 30x^2 - 54x + 4) \][/tex]
[tex]\[ f'(x) = -6x^2 + 60x - 54 \][/tex]
### Step 2: Find Critical Points
To find the critical points, we set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -6x^2 + 60x - 54 = 0 \][/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]\( a = -6 \)[/tex], [tex]\( b = 60 \)[/tex], and [tex]\( c = -54 \)[/tex].
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-60 \pm \sqrt{60^2 - 4(-6)(-54)}}{2(-6)} \][/tex]
[tex]\[ x = \frac{-60 \pm \sqrt{3600 - 1296}}{-12} \][/tex]
[tex]\[ x = \frac{-60 \pm \sqrt{2304}}{-12} \][/tex]
[tex]\[ x = \frac{-60 \pm 48}{-12} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-60 + 48}{-12} = 1 \][/tex]
[tex]\[ x = \frac{-60 - 48}{-12} = 9 \][/tex]
So, the critical points are [tex]\( x = 1 \)[/tex] and [tex]\( x = 9 \)[/tex].
### Step 3: Find the Second Derivative
To classify these critical points, we find the second derivative, [tex]\( f''(x) \)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx} (-6x^2 + 60x - 54) \][/tex]
[tex]\[ f''(x) = -12x + 60 \][/tex]
### Step 4: Apply the Second Derivative Test
We evaluate the second derivative at each of the critical points:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f''(1) = -12(1) + 60 = 48 \][/tex]
Since [tex]\( f''(1) > 0 \)[/tex], the function has a local minimum at [tex]\( x = 1 \)[/tex].
For [tex]\( x = 9 \)[/tex]:
[tex]\[ f''(9) = -12(9) + 60 = -48 \][/tex]
Since [tex]\( f''(9) < 0 \)[/tex], the function has a local maximum at [tex]\( x = 9 \)[/tex].
### Step 5: Evaluate the Function at the Critical Points
Now we calculate the function values at these critical points:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -2(1)^3 + 30(1)^2 - 54(1) + 4 \][/tex]
[tex]\[ f(1) = -2 + 30 - 54 + 4 \][/tex]
[tex]\[ f(1) = -22 \][/tex]
For [tex]\( x = 9 \)[/tex]:
[tex]\[ f(9) = -2(9)^3 + 30(9)^2 - 54(9) + 4 \][/tex]
[tex]\[ f(9) = -2(729) + 30(81) - 54(9) + 4 \][/tex]
[tex]\[ f(9) = -1458 + 2430 - 486 + 4 \][/tex]
[tex]\[ f(9) = 490 \][/tex]
### Final Answer
So, the function has a local minimum at [tex]\( x = 1 \)[/tex] with value [tex]\( -22 \)[/tex] and a local maximum at [tex]\( x = 9 \)[/tex] with value [tex]\( 490 \)[/tex].
- Local minimum: [tex]\( x = 1 \)[/tex], value = [tex]\( -22 \)[/tex]
- Local maximum: [tex]\( x = 9 \)[/tex], value = [tex]\( 490 \)[/tex]
### Step 1: Find the First Derivative
The first derivative of the function [tex]\( f(x) \)[/tex], denoted as [tex]\( f'(x) \)[/tex], is found using the power rule:
[tex]\[ f'(x) = \frac{d}{dx} (-2x^3 + 30x^2 - 54x + 4) \][/tex]
[tex]\[ f'(x) = -6x^2 + 60x - 54 \][/tex]
### Step 2: Find Critical Points
To find the critical points, we set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -6x^2 + 60x - 54 = 0 \][/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]\( a = -6 \)[/tex], [tex]\( b = 60 \)[/tex], and [tex]\( c = -54 \)[/tex].
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-60 \pm \sqrt{60^2 - 4(-6)(-54)}}{2(-6)} \][/tex]
[tex]\[ x = \frac{-60 \pm \sqrt{3600 - 1296}}{-12} \][/tex]
[tex]\[ x = \frac{-60 \pm \sqrt{2304}}{-12} \][/tex]
[tex]\[ x = \frac{-60 \pm 48}{-12} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-60 + 48}{-12} = 1 \][/tex]
[tex]\[ x = \frac{-60 - 48}{-12} = 9 \][/tex]
So, the critical points are [tex]\( x = 1 \)[/tex] and [tex]\( x = 9 \)[/tex].
### Step 3: Find the Second Derivative
To classify these critical points, we find the second derivative, [tex]\( f''(x) \)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx} (-6x^2 + 60x - 54) \][/tex]
[tex]\[ f''(x) = -12x + 60 \][/tex]
### Step 4: Apply the Second Derivative Test
We evaluate the second derivative at each of the critical points:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f''(1) = -12(1) + 60 = 48 \][/tex]
Since [tex]\( f''(1) > 0 \)[/tex], the function has a local minimum at [tex]\( x = 1 \)[/tex].
For [tex]\( x = 9 \)[/tex]:
[tex]\[ f''(9) = -12(9) + 60 = -48 \][/tex]
Since [tex]\( f''(9) < 0 \)[/tex], the function has a local maximum at [tex]\( x = 9 \)[/tex].
### Step 5: Evaluate the Function at the Critical Points
Now we calculate the function values at these critical points:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -2(1)^3 + 30(1)^2 - 54(1) + 4 \][/tex]
[tex]\[ f(1) = -2 + 30 - 54 + 4 \][/tex]
[tex]\[ f(1) = -22 \][/tex]
For [tex]\( x = 9 \)[/tex]:
[tex]\[ f(9) = -2(9)^3 + 30(9)^2 - 54(9) + 4 \][/tex]
[tex]\[ f(9) = -2(729) + 30(81) - 54(9) + 4 \][/tex]
[tex]\[ f(9) = -1458 + 2430 - 486 + 4 \][/tex]
[tex]\[ f(9) = 490 \][/tex]
### Final Answer
So, the function has a local minimum at [tex]\( x = 1 \)[/tex] with value [tex]\( -22 \)[/tex] and a local maximum at [tex]\( x = 9 \)[/tex] with value [tex]\( 490 \)[/tex].
- Local minimum: [tex]\( x = 1 \)[/tex], value = [tex]\( -22 \)[/tex]
- Local maximum: [tex]\( x = 9 \)[/tex], value = [tex]\( 490 \)[/tex]
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