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Magazine Chapter 2: Problem 27
Find the relative extrema of each function, if they exist. List each extremum along with the \(x\) -value at which it occurs. Then sketch a graph of the function. $$G(x)=\frac{-8}{x^{2}+1}$$
Short Answer
Expert verified
The relative extremum for \[G(x) = \frac{-8}{x^2 + 1}\] is a minimum at \[x = 0\] with value \[-8\].
Step by step solution
01
Find the first derivative of the function
First, find the first derivative of the function to identify critical points. For the function \[G(x) = \frac{-8}{x^2 + 1}\]use the quotient rule: \[G'(x) = \frac{d}{dx}\left( \frac{-8}{x^2 + 1} \right)\]Let \[u = -8\]and \[v = x^2 + 1\].
02
Apply the quotient rule
Using the quotient rule, which is \[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2},\]we get \[u' = 0\]and \[v' = 2x\]. Substitute these into the quotient rule formula: \[G'(x) = \frac{(0)(x^2 + 1) - (-8)(2x)}{(x^2 + 1)^2} = \frac{16x}{(x^2 + 1)^2}.\]
03
Find critical points by setting the first derivative to zero
To find critical points, set \[G'(x) = 0\]:\[\frac{16x}{(x^2 + 1)^2} = 0\]The numerator must be zero: \[16x = 0 \implies x = 0.\]
04
Find the second derivative to identify concavity
Find the second derivative \[G''(x)\] to determine the concavity. Using the product rule on \[G'(x) = \frac{16x}{(x^2 + 1)^2},\] let \[u = 16x\]and \[v = (x^2 + 1)^{-2},\] we get:\[u' = 16\]and \[v' = -2(x^2 + 1)^{-3}(2x) = -4x(x^2 + 1)^{-3}.\]Using the product rule \[(uv')' = u'v + uv',\] we have:\[G''(x) = 16(x^2 + 1)^{-2} + 16x(-4x(x^2 + 1)^{-3})\]\[= 16(x^2 + 1)^{-2} - 64x^2(x^2 + 1)^{-3}\]\[= 16(x^2 + 1)^{-3}((x^2 + 1) - 4x^2).\]
05
Simplify the second derivative expression
Simplify the second derivative:\[G''(x) = 16(x^2 + 1)^{-3}(1 - 3x^2)\]. Evaluate this at the critical point \[x = 0\]:\[G''(0) = 16(1)^{-3}(1 - 0) = 16 > 0,\]indicating a local minimum.
06
Identify and list relative extremum
Since \[G''(0) > 0\], the critical point \[x = 0\] is a local minimum for \[G(x)\]. The minimum value when \[x = 0\] is \[G(0) = \frac{-8}{0^2 + 1} = -8\]. Thus, the relative extremum is a minimum at \[x = 0\] with value \[-8\].
07
Sketch the graph
The graph of \[G(x) = \frac{-8}{x^2 + 1}\] is a downward-opening curve that approaches \[y = 0\] as \[|x|\] increases. The local minimum at \[x = 0\] corresponds to \[y = -8\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
first derivative
To find the relative extrema of a function, we first need the first derivative. The first derivative tells us the rate of change of the function.
For the function \(G(x) = \frac{-8}{x^2 + 1}\), we use the quotient rule for differentiation. The quotient rule is: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}.\]
Here, \(u = -8\) and \(v = x^2 + 1\). Applying the rule, we get:
\[ G'(x) = \frac{(0)(x^2 + 1) - (-8)(2x)}{(x^2 + 1)^2} = \frac{16x}{(x^2 + 1)^2}. \]Identifying the critical points is our next step.
For the function \(G(x) = \frac{-8}{x^2 + 1}\), we use the quotient rule for differentiation. The quotient rule is: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}.\]
Here, \(u = -8\) and \(v = x^2 + 1\). Applying the rule, we get:
- \(u' = 0\)
- \(v' = 2x\)
\[ G'(x) = \frac{(0)(x^2 + 1) - (-8)(2x)}{(x^2 + 1)^2} = \frac{16x}{(x^2 + 1)^2}. \]Identifying the critical points is our next step.
critical points
Critical points occur where the first derivative is zero or undefined. These points are where the function could have a relative extremum (a local maximum or minimum).
For our function, set the first derivative to zero:
\[ \frac{16x}{(x^2 + 1)^2} = 0. \]
The numerator, \(16x\), must be zero:
\[ 16x = 0 \implies x = 0. \]
The critical point is at \(x = 0\). To determine the nature of the extremum, we need to examine the second derivative.
For our function, set the first derivative to zero:
\[ \frac{16x}{(x^2 + 1)^2} = 0. \]
The numerator, \(16x\), must be zero:
\[ 16x = 0 \implies x = 0. \]
The critical point is at \(x = 0\). To determine the nature of the extremum, we need to examine the second derivative.
concavity
Concavity helps to identify whether a critical point is a local minimum, maximum, or a point of inflection.
By finding the second derivative, we can learn about the concavity of the graph. If the second derivative at a point is positive, the function is concave up (shaped like a cup) and has a local minimum. If negative, it is concave down (shaped like a cap) and has a local maximum.
By finding the second derivative, we can learn about the concavity of the graph. If the second derivative at a point is positive, the function is concave up (shaped like a cup) and has a local minimum. If negative, it is concave down (shaped like a cap) and has a local maximum.
second derivative
We use the product rule for the second derivative for our function.
We find:
\[ G''(x) = 16(x^2 + 1)^{-2} - 64x^2(x^2 + 1)^{-3} \] Simplifying, we get:
\[ G''(x) = 16(x^2 + 1)^{-3}(1 - 3x^2) \].
Evaluate the second derivative at \(x = 0\):
\[ G''(0) = 16(1)^{-3}(1 - 0) = 16 > 0. \]This indicates the critical point \(x = 0\) is a local minimum.
- \(u = 16x\)
- \(v = (x^2 + 1)^{-2}\)
- \(u' = 16\)
- \(v' = -4x(x^2 + 1)^{-3}\)
We find:
\[ G''(x) = 16(x^2 + 1)^{-2} - 64x^2(x^2 + 1)^{-3} \] Simplifying, we get:
\[ G''(x) = 16(x^2 + 1)^{-3}(1 - 3x^2) \].
Evaluate the second derivative at \(x = 0\):
\[ G''(0) = 16(1)^{-3}(1 - 0) = 16 > 0. \]This indicates the critical point \(x = 0\) is a local minimum.
local minimum
A local minimum is a point where the function reaches a lower value than at any nearby points.
To conclude, the second derivative test confirmed that \(x = 0\) is a local minimum. At this point, the function value is:
\[ G(0) = \frac{-8}{0^2 + 1} = -8. \]
Thus, the function \( G(x) \) has a local minimum at \( x = 0 \) with a value of -8.
To conclude, the second derivative test confirmed that \(x = 0\) is a local minimum. At this point, the function value is:
\[ G(0) = \frac{-8}{0^2 + 1} = -8. \]
Thus, the function \( G(x) \) has a local minimum at \( x = 0 \) with a value of -8.
