Find all relative extrema of the function. Use the second derivative test where applicable. (If an answer does not exist, enter DNE.)
[tex]f(x)=x+\frac{25}{x}[/tex]
relative maximum (x, y) = (
relative minimum (x, y) = (
Answer (1)
Find the first derivative: f ′ ( x ) = 1 − x 2 25 .
Find the critical points by setting f ′ ( x ) = 0 : x = ± 5 .
Find the second derivative: f ′′ ( x ) = x 3 50 .
Apply the second derivative test: 0"> f ′′ ( 5 ) > 0 indicates a relative minimum at ( 5 , 10 ) , and f ′′ ( − 5 ) < 0 indicates a relative maximum at ( − 5 , − 10 ) .
relative maximum ( − 5 , − 10 ) , relative minimum ( 5 , 10 )
Explanation
Problem Analysis We are given the function f ( x ) = x + x 25 and we need to find its relative extrema using the second derivative test where applicable.
Find the First Derivative First, we find the first derivative of the function:
First Derivative f ′ ( x ) = 1 − x 2 25
Find Critical Points Next, we find the critical points by setting the first derivative equal to zero and solving for x :
Critical Points 1 − x 2 25 = 0 ⇒ x 2 = 25 ⇒ x = ± 5
Find the Second Derivative Now, we find the second derivative of the function:
Second Derivative f ′′ ( x ) = x 3 50
Second Derivative Test We use the second derivative test to determine if each critical point is a relative maximum or minimum. If 0"> f ′′ ( c ) > 0 , then f ( c ) is a relative minimum. If f ′′ ( c ) < 0 , then f ( c ) is a relative maximum.
Relative Minimum For x = 5 , 0"> f ′′ ( 5 ) = 5 3 50 = 125 50 = 5 2 > 0 , so f ( 5 ) is a relative minimum. f ( 5 ) = 5 + 5 25 = 5 + 5 = 10 . Thus, there is a relative minimum at ( 5 , 10 ) .
Relative Maximum For x = − 5 , f ′′ ( − 5 ) = ( − 5 ) 3 50 = − 125 50 = − 5 2 < 0 , so f ( − 5 ) is a relative maximum. f ( − 5 ) = − 5 + − 5 25 = − 5 − 5 = − 10 . Thus, there is a relative maximum at ( − 5 , − 10 ) .
Final Answer Therefore, the relative maximum is at ( − 5 , − 10 ) and the relative minimum is at ( 5 , 10 ) .
Examples
Understanding extrema is crucial in optimization problems. For instance, if you're designing a container to minimize surface area for a given volume, finding the minimum of a surface area function helps determine the most efficient dimensions. Similarly, in economics, businesses use calculus to maximize profit or minimize cost by finding extrema of cost and revenue functions. These optimization techniques are fundamental in engineering, economics, and various other fields.
