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To find the local minimum and maximum of the function [tex]\( f(x) = -2x^3 + 42x^2 - 240x + 8 \)[/tex], we need to follow several mathematical steps.
### Step 1: Calculate the first derivative of [tex]\( f(x) \)[/tex]
The first derivative [tex]\( f'(x) \)[/tex] helps us find the critical points of the function.
[tex]\[ f'(x) = \frac{d}{dx} \left( -2x^3 + 42x^2 - 240x + 8 \right) \][/tex]
[tex]\[ f'(x) = -6x^2 + 84x - 240 \][/tex]
### Step 2: Find the critical points by setting the first derivative to zero
We solve the equation [tex]\( f'(x) = 0 \)[/tex] to find the critical points.
[tex]\[ -6x^2 + 84x - 240 = 0 \][/tex]
This is a quadratic equation, and solving it gives us two values for [tex]\( x \)[/tex]:
[tex]\[ x = 4 \][/tex]
[tex]\[ x = 10 \][/tex]
### Step 3: Calculate the second derivative of [tex]\( f(x) \)[/tex]
The second derivative [tex]\( f''(x) \)[/tex] helps us determine the concavity at the critical points.
[tex]\[ f''(x) = \frac{d}{dx} \left( -6x^2 + 84x - 240 \right) \][/tex]
[tex]\[ f''(x) = -12x + 84 \][/tex]
### Step 4: Determine the concavity at each critical point
We evaluate the second derivative at [tex]\( x = 4 \)[/tex] and [tex]\( x = 10 \)[/tex].
For [tex]\( x = 4 \)[/tex]:
[tex]\[ f''(4) = -12(4) + 84 = -48 + 84 = 36 \][/tex]
Since [tex]\( f''(4) > 0 \)[/tex], the function is concave up at [tex]\( x = 4 \)[/tex], indicating a local minimum.
For [tex]\( x = 10 \)[/tex]:
[tex]\[ f''(10) = -12(10) + 84 = -120 + 84 = -36 \][/tex]
Since [tex]\( f''(10) < 0 \)[/tex], the function is concave down at [tex]\( x = 10 \)[/tex], indicating a local maximum.
### Step 5: Determine the function values at the critical points
To find the local minimum and maximum values, we substitute [tex]\( x = 4 \)[/tex] and [tex]\( x = 10 \)[/tex] back into the original function [tex]\( f(x) \)[/tex].
For [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = -2(4)^3 + 42(4)^2 - 240(4) + 8 \][/tex]
[tex]\[ f(4) = -128 + 672 - 960 + 8 = -408 \][/tex]
For [tex]\( x = 10 \)[/tex]:
[tex]\[ f(10) = -2(10)^3 + 42(10)^2 - 240(10) + 8 \][/tex]
[tex]\[ f(10) = -2000 + 4200 - 2400 + 8 = -192 \][/tex]
### Conclusion:
- The function has a local minimum at [tex]\( x = 4 \)[/tex] with a value of [tex]\( -408 \)[/tex].
- The function has a local maximum at [tex]\( x = 10 \)[/tex] with a value of [tex]\( -192 \)[/tex].
Thus, the completed statement becomes:
The function [tex]\( f(x) = -2 x^3 + 42 x^2 - 240 x + 8 \)[/tex] has one local minimum and one local maximum.
This function has a local minimum at [tex]\( x = 4 \)[/tex] with value [tex]\( -408 \)[/tex] and a local maximum at [tex]\( x = 10 \)[/tex] with value [tex]\( -192 \)[/tex].
### Step 1: Calculate the first derivative of [tex]\( f(x) \)[/tex]
The first derivative [tex]\( f'(x) \)[/tex] helps us find the critical points of the function.
[tex]\[ f'(x) = \frac{d}{dx} \left( -2x^3 + 42x^2 - 240x + 8 \right) \][/tex]
[tex]\[ f'(x) = -6x^2 + 84x - 240 \][/tex]
### Step 2: Find the critical points by setting the first derivative to zero
We solve the equation [tex]\( f'(x) = 0 \)[/tex] to find the critical points.
[tex]\[ -6x^2 + 84x - 240 = 0 \][/tex]
This is a quadratic equation, and solving it gives us two values for [tex]\( x \)[/tex]:
[tex]\[ x = 4 \][/tex]
[tex]\[ x = 10 \][/tex]
### Step 3: Calculate the second derivative of [tex]\( f(x) \)[/tex]
The second derivative [tex]\( f''(x) \)[/tex] helps us determine the concavity at the critical points.
[tex]\[ f''(x) = \frac{d}{dx} \left( -6x^2 + 84x - 240 \right) \][/tex]
[tex]\[ f''(x) = -12x + 84 \][/tex]
### Step 4: Determine the concavity at each critical point
We evaluate the second derivative at [tex]\( x = 4 \)[/tex] and [tex]\( x = 10 \)[/tex].
For [tex]\( x = 4 \)[/tex]:
[tex]\[ f''(4) = -12(4) + 84 = -48 + 84 = 36 \][/tex]
Since [tex]\( f''(4) > 0 \)[/tex], the function is concave up at [tex]\( x = 4 \)[/tex], indicating a local minimum.
For [tex]\( x = 10 \)[/tex]:
[tex]\[ f''(10) = -12(10) + 84 = -120 + 84 = -36 \][/tex]
Since [tex]\( f''(10) < 0 \)[/tex], the function is concave down at [tex]\( x = 10 \)[/tex], indicating a local maximum.
### Step 5: Determine the function values at the critical points
To find the local minimum and maximum values, we substitute [tex]\( x = 4 \)[/tex] and [tex]\( x = 10 \)[/tex] back into the original function [tex]\( f(x) \)[/tex].
For [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = -2(4)^3 + 42(4)^2 - 240(4) + 8 \][/tex]
[tex]\[ f(4) = -128 + 672 - 960 + 8 = -408 \][/tex]
For [tex]\( x = 10 \)[/tex]:
[tex]\[ f(10) = -2(10)^3 + 42(10)^2 - 240(10) + 8 \][/tex]
[tex]\[ f(10) = -2000 + 4200 - 2400 + 8 = -192 \][/tex]
### Conclusion:
- The function has a local minimum at [tex]\( x = 4 \)[/tex] with a value of [tex]\( -408 \)[/tex].
- The function has a local maximum at [tex]\( x = 10 \)[/tex] with a value of [tex]\( -192 \)[/tex].
Thus, the completed statement becomes:
The function [tex]\( f(x) = -2 x^3 + 42 x^2 - 240 x + 8 \)[/tex] has one local minimum and one local maximum.
This function has a local minimum at [tex]\( x = 4 \)[/tex] with value [tex]\( -408 \)[/tex] and a local maximum at [tex]\( x = 10 \)[/tex] with value [tex]\( -192 \)[/tex].
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