IDNLearn.com is the perfect place to get detailed and accurate answers to your questions. Our experts provide timely and accurate responses to help you navigate any topic or issue with confidence.
Sagot :
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].
Your participation is crucial to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. Discover the answers you need at IDNLearn.com. Thanks for visiting, and come back soon for more valuable insights.
