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^2+5 x-5[/tex]
relative maximum (x, y) = (
relative minimum (x, y) = (
Answer (1)
Find the first derivative: f ′ ( x ) = 2 x + 5 .
Find the critical point by solving f ′ ( x ) = 0 , which gives x = − 2 5 .
Find the second derivative: f ′′ ( x ) = 2 .
Apply the second derivative test: since 0"> f ′′ ( − 2 5 ) = 2 > 0 , there is a relative minimum at x = − 2 5 , and its y -coordinate is f ( − 2 5 ) = − 4 45 . There is no relative maximum.
The relative minimum is at ( − 2 5 , − 4 45 ) .
Explanation
Problem Analysis We are given the function f ( x ) = x 2 + 5 x − 5 and asked to find its relative extrema using the second derivative test.
Finding the First Derivative First, we find the first derivative of the function: f ′ ( x ) = 2 x + 5
Finding Critical Points Next, we find the critical points by setting the first derivative equal to zero and solving for x :
2 x + 5 = 0 2 x = − 5 x = − 2 5
Finding the Second Derivative Now, we find the second derivative of the function: f ′′ ( x ) = 2
Applying the Second Derivative Test We apply the second derivative test. Since 0"> f ′′ ( − 2 5 ) = 2 > 0 , the function has a relative minimum at x = − 2 5 .
Finding the y-coordinate To find the y -coordinate of the relative minimum, we plug x = − 2 5 into the original function: f ( − 2 5 ) = ( − 2 5 ) 2 + 5 ( − 2 5 ) − 5 f ( − 2 5 ) = 4 25 − 2 25 − 5 f ( − 2 5 ) = 4 25 − 4 50 − 4 20 f ( − 2 5 ) = 4 25 − 50 − 20 f ( − 2 5 ) = 4 − 45 So, the relative minimum is at ( − 2 5 , − 4 45 ) .
Conclusion Since the second derivative is always positive, there is no relative maximum.
Examples
Understanding extrema is crucial in optimization problems, such as maximizing profit or minimizing cost in business. For instance, a company might use calculus to find the production level that minimizes the average cost per unit, helping them make informed decisions about their operations and pricing strategies.
