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To find the local maximum and minimum of the function [tex]\( f(x)=7x+3x^{-1} \)[/tex], we will go through the following steps:
### Step 1: Find the Derivative
First, calculate the first derivative of the function to find the critical points.
The function given is:
[tex]\[ f(x) = 7x + 3x^{-1} \][/tex]
Calculate the first derivative:
[tex]\[ f'(x) = \frac{d}{dx}(7x + 3x^{-1}) \][/tex]
[tex]\[ f'(x) = 7 - 3x^{-2} \][/tex]
### Step 2: Set the First Derivative to Zero
We set the first derivative equal to zero to find the critical points:
[tex]\[ 7 - 3x^{-2} = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 7 = 3x^{-2} \][/tex]
[tex]\[ 7 = \frac{3}{x^2} \][/tex]
[tex]\[ 7x^2 = 3 \][/tex]
[tex]\[ x^2 = \frac{3}{7} \][/tex]
[tex]\[ x = \pm \sqrt{\frac{3}{7}} \][/tex]
[tex]\[ x = \pm \frac{\sqrt{21}}{7} \][/tex]
So, the critical points are [tex]\( x = \frac{\sqrt{21}}{7} \)[/tex] and [tex]\( x = -\frac{\sqrt{21}}{7} \)[/tex].
### Step 3: Second Derivative Test
Next, calculate the second derivative to classify these critical points:
[tex]\[ f''(x) = \frac{d}{dx} (7 - 3x^{-2}) \][/tex]
[tex]\[ f''(x) = 6x^{-3} \][/tex]
[tex]\[ f''(x) = \frac{6}{x^3} \][/tex]
### Step 4: Evaluate the Second Derivative at Critical Points
Evaluate the second derivative at each critical point to determine if it is a local maximum or minimum.
For [tex]\( x = \frac{\sqrt{21}}{7} \)[/tex]:
[tex]\[ f''\left(\frac{\sqrt{21}}{7}\right) = \frac{6}{\left(\frac{\sqrt{21}}{7}\right)^3} = \frac{6 \cdot 7^3}{21^{3/2}} = \frac{6 \cdot 343}{21\sqrt{21}} = \frac{2058}{21\sqrt{21}} > 0 \][/tex]
Since [tex]\( f''\left(\frac{\sqrt{21}}{7}\right) > 0 \)[/tex], [tex]\( x = \frac{\sqrt{21}}{7} \)[/tex] is a local minimum.
For [tex]\( x = -\frac{\sqrt{21}}{7} \)[/tex]:
[tex]\[ f''\left(-\frac{\sqrt{21}}{7}\right) = \frac{6}{\left(-\frac{\sqrt{21}}{7}\right)^3} = -\frac{6 \cdot 343}{21^{3/2}} = -\frac{2058}{21\sqrt{21}} < 0 \][/tex]
Since [tex]\( f''\left(-\frac{\sqrt{21}}{7}\right) < 0 \)[/tex], [tex]\( x = -\frac{\sqrt{21}}{7} \)[/tex] is a local maximum.
### Step 5: Find Function Values at Critical Points
Calculate the function values at these critical points.
For [tex]\( x = \frac{\sqrt{21}}{7} \)[/tex]:
[tex]\[ f\left(\frac{\sqrt{21}}{7}\right) = 7\left(\frac{\sqrt{21}}{7}\right) + 3\left(\frac{7}{\sqrt{21}}\right) \][/tex]
[tex]\[ f\left(\frac{\sqrt{21}}{7}\right) = \sqrt{21} + \frac{21}{\sqrt{21}} = 2\sqrt{21} \][/tex]
For [tex]\( x = -\frac{\sqrt{21}}{7} \)[/tex]:
[tex]\[ f\left(-\frac{\sqrt{21}}{7}\right) = 7\left(-\frac{\sqrt{21}}{7}\right) + 3\left(-\frac{7}{\sqrt{21}}\right) \][/tex]
[tex]\[ f\left(-\frac{\sqrt{21}}{7}\right) = -\sqrt{21} - \frac{21}{\sqrt{21}} = -2\sqrt{21} \][/tex]
### Final Answer:
- The function [tex]\( f(x) \)[/tex] has a local maximum at [tex]\( x = -\frac{\sqrt{21}}{7} \)[/tex] with value [tex]\( -2\sqrt{21} \)[/tex].
- The function [tex]\( f(x) \)[/tex] has a local minimum at [tex]\( x = \frac{\sqrt{21}}{7} \)[/tex] with value [tex]\( 2\sqrt{21} \)[/tex].
### Step 1: Find the Derivative
First, calculate the first derivative of the function to find the critical points.
The function given is:
[tex]\[ f(x) = 7x + 3x^{-1} \][/tex]
Calculate the first derivative:
[tex]\[ f'(x) = \frac{d}{dx}(7x + 3x^{-1}) \][/tex]
[tex]\[ f'(x) = 7 - 3x^{-2} \][/tex]
### Step 2: Set the First Derivative to Zero
We set the first derivative equal to zero to find the critical points:
[tex]\[ 7 - 3x^{-2} = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 7 = 3x^{-2} \][/tex]
[tex]\[ 7 = \frac{3}{x^2} \][/tex]
[tex]\[ 7x^2 = 3 \][/tex]
[tex]\[ x^2 = \frac{3}{7} \][/tex]
[tex]\[ x = \pm \sqrt{\frac{3}{7}} \][/tex]
[tex]\[ x = \pm \frac{\sqrt{21}}{7} \][/tex]
So, the critical points are [tex]\( x = \frac{\sqrt{21}}{7} \)[/tex] and [tex]\( x = -\frac{\sqrt{21}}{7} \)[/tex].
### Step 3: Second Derivative Test
Next, calculate the second derivative to classify these critical points:
[tex]\[ f''(x) = \frac{d}{dx} (7 - 3x^{-2}) \][/tex]
[tex]\[ f''(x) = 6x^{-3} \][/tex]
[tex]\[ f''(x) = \frac{6}{x^3} \][/tex]
### Step 4: Evaluate the Second Derivative at Critical Points
Evaluate the second derivative at each critical point to determine if it is a local maximum or minimum.
For [tex]\( x = \frac{\sqrt{21}}{7} \)[/tex]:
[tex]\[ f''\left(\frac{\sqrt{21}}{7}\right) = \frac{6}{\left(\frac{\sqrt{21}}{7}\right)^3} = \frac{6 \cdot 7^3}{21^{3/2}} = \frac{6 \cdot 343}{21\sqrt{21}} = \frac{2058}{21\sqrt{21}} > 0 \][/tex]
Since [tex]\( f''\left(\frac{\sqrt{21}}{7}\right) > 0 \)[/tex], [tex]\( x = \frac{\sqrt{21}}{7} \)[/tex] is a local minimum.
For [tex]\( x = -\frac{\sqrt{21}}{7} \)[/tex]:
[tex]\[ f''\left(-\frac{\sqrt{21}}{7}\right) = \frac{6}{\left(-\frac{\sqrt{21}}{7}\right)^3} = -\frac{6 \cdot 343}{21^{3/2}} = -\frac{2058}{21\sqrt{21}} < 0 \][/tex]
Since [tex]\( f''\left(-\frac{\sqrt{21}}{7}\right) < 0 \)[/tex], [tex]\( x = -\frac{\sqrt{21}}{7} \)[/tex] is a local maximum.
### Step 5: Find Function Values at Critical Points
Calculate the function values at these critical points.
For [tex]\( x = \frac{\sqrt{21}}{7} \)[/tex]:
[tex]\[ f\left(\frac{\sqrt{21}}{7}\right) = 7\left(\frac{\sqrt{21}}{7}\right) + 3\left(\frac{7}{\sqrt{21}}\right) \][/tex]
[tex]\[ f\left(\frac{\sqrt{21}}{7}\right) = \sqrt{21} + \frac{21}{\sqrt{21}} = 2\sqrt{21} \][/tex]
For [tex]\( x = -\frac{\sqrt{21}}{7} \)[/tex]:
[tex]\[ f\left(-\frac{\sqrt{21}}{7}\right) = 7\left(-\frac{\sqrt{21}}{7}\right) + 3\left(-\frac{7}{\sqrt{21}}\right) \][/tex]
[tex]\[ f\left(-\frac{\sqrt{21}}{7}\right) = -\sqrt{21} - \frac{21}{\sqrt{21}} = -2\sqrt{21} \][/tex]
### Final Answer:
- The function [tex]\( f(x) \)[/tex] has a local maximum at [tex]\( x = -\frac{\sqrt{21}}{7} \)[/tex] with value [tex]\( -2\sqrt{21} \)[/tex].
- The function [tex]\( f(x) \)[/tex] has a local minimum at [tex]\( x = \frac{\sqrt{21}}{7} \)[/tex] with value [tex]\( 2\sqrt{21} \)[/tex].
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