Find the absolute extrema of the function on the closed interval.
Consider the following function and closed interval:
[tex]g(x)=\frac{2 x^2}{x-2}, \quad[-2,1][/tex]
Find [tex]g^{\prime}(x)[/tex].
[tex]g^{\prime}(x)=\square[/tex]
Find the critical numbers of [tex]g[/tex] in [tex](-2,1)[/tex] and evaluate [tex]g[/tex] at each critical number.
[tex](x, y)=(\square)[/tex]
Evaluate [tex]g[/tex] at each endpoint of [tex][-2,1][/tex].
Left endpoint: [tex](x, y)=[/tex] ([tex]\square[/tex])
Right endpoint: [tex](x, y)=[/tex] ([tex]\square[/tex])
Find the absolute extrema of the function on the closed interval [tex][-2,1][/tex].
Minima:
Smaller [tex]x[/tex]-value: [tex](x, y)=([/tex] [tex]\square[/tex])
Larger [tex]x[/tex]-value: [tex](x, y)=([/tex] [tex]\square[/tex])
Maximum:
[tex](x, y)=(\square)[/tex]
Answer (2)
Find the derivative of the function g ( x ) = x − 2 2 x 2 using the quotient rule: g ′ ( x ) = ( x − 2 ) 2 2 x ( x − 4 ) .
Determine the critical points by setting g ′ ( x ) = 0 , which gives x = 0 and x = 4 . Only x = 0 lies in the interval [ − 2 , 1 ] .
Evaluate g ( x ) at the critical point x = 0 and the endpoints x = − 2 and x = 1 : g ( 0 ) = 0 , g ( − 2 ) = − 2 , and g ( 1 ) = − 2 .
Identify the absolute minimum and maximum values: the absolute minimum is − 2 at x = − 2 and x = 1 , and the absolute maximum is 0 at x = 0 . Thus, the absolute extrema are ( − 2 , − 2 ) and ( 1 , − 2 ) for minima and ( 0 , 0 ) for the maximum.
Explanation
Problem Analysis We are given the function g ( x ) = x − 2 2 x 2 and the interval [ − 2 , 1 ] . Our goal is to find the absolute maximum and minimum values of g ( x ) on this interval. To do this, we will first find the critical points of g ( x ) by finding where its derivative is equal to zero or undefined. Then, we will evaluate g ( x ) at the critical points and the endpoints of the interval to determine the absolute extrema.
Finding the Derivative First, we need to find the derivative of g ( x ) . Using the quotient rule, we have:
g ′ ( x ) = ( x − 2 ) 2 ( x − 2 ) ( 4 x ) − ( 2 x 2 ) ( 1 ) = ( x − 2 ) 2 4 x 2 − 8 x − 2 x 2 = ( x − 2 ) 2 2 x 2 − 8 x = ( x − 2 ) 2 2 x ( x − 4 ) So, g ′ ( x ) = ( x − 2 ) 2 2 x ( x − 4 ) .
Finding Critical Points Next, we find the critical points by setting g ′ ( x ) = 0 . This occurs when the numerator is zero, i.e., 2 x ( x − 4 ) = 0 . This gives us x = 0 and x = 4 . Also, g ′ ( x ) is undefined when the denominator is zero, i.e., ( x − 2 ) 2 = 0 , which gives x = 2 . However, x = 2 is not in the domain of g ( x ) , so it is not a critical point.
Now, we need to check which critical points are in the interval ( − 2 , 1 ) . The critical point x = 0 is in the interval ( − 2 , 1 ) , but x = 4 is not. Thus, we only need to consider x = 0 .
We evaluate g ( x ) at x = 0 :
g ( 0 ) = 0 − 2 2 ( 0 ) 2 = 0 So, we have the critical point ( 0 , 0 ) .
Evaluating at Endpoints Now, we evaluate g ( x ) at the endpoints of the interval [ − 2 , 1 ] .
Left endpoint: x = − 2
g ( − 2 ) = − 2 − 2 2 ( − 2 ) 2 = − 4 2 ( 4 ) = − 2 So, we have the point ( − 2 , − 2 ) .
Right endpoint: x = 1
g ( 1 ) = 1 − 2 2 ( 1 ) 2 = − 1 2 = − 2 So, we have the point ( 1 , − 2 ) .
Finding Absolute Extrema Finally, we compare the values of g ( x ) at the critical point and the endpoints:
g ( 0 ) = 0 g ( − 2 ) = − 2 g ( 1 ) = − 2
The minimum value is − 2 , which occurs at x = − 2 and x = 1 . The maximum value is 0 , which occurs at x = 0 .
Therefore, the absolute minimum is − 2 and the absolute maximum is 0 .
Final Answer The absolute minima occur at ( − 2 , − 2 ) and ( 1 , − 2 ) . The absolute maximum occurs at ( 0 , 0 ) .
Examples
Imagine you're designing a bridge and need to minimize the stress on a particular support beam. The stress can be modeled by a function similar to the one in this problem. By finding the absolute extrema of the stress function over a given interval, you can determine the maximum and minimum stress values, ensuring the bridge's structural integrity and safety. This helps engineers make informed decisions about material selection and design parameters.
The absolute minima of the function g ( x ) occur at ( − 2 , − 2 ) and ( 1 , − 2 ) , while the absolute maximum occurs at ( 0 , 0 ) .
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