How do you find all local maximum and minimum points given #y=x^2-98x+4#?

Question

Aurora Alvarado · Accepted Answer

The minimum point is at # (49, -2397) # [Ans]

#y= x^2-98x+4 # . At turning point , #dy/dx=0# # dy/dx= 2x-98 :. 2x -98=0 or 2x =98 or x =49 # At #x=49 ; y = 49^2 -98*49+4 or y= -2397# So turrning point is at # (49, -2397) #. #(d^2y)/dx^2= 2 # . To distinguish maximum ar minimum point we know if #(d^2y)/dx^2 > 0 # then the point must be a minimum. Here #(d^2y)/dx^2 > 0 # , so the point # (49, -2397) # is minimum. The minimum point is at # (49, -2397) # [Ans]

William Abdalla · Answer

undefined To find all local maximum and minimum points of the function y = x^2 - 98x + 4, you need to follow these steps:

1. Compute the derivative of the function.
2. Find the critical points by setting the derivative equal to zero and solving for x.
3. Determine the nature of each critical point using the second derivative test.
4. Identify the local maximum and minimum points based on the nature of the critical points.

Let's go through these steps:

1. The derivative of the function y = x^2 - 98x + 4 is y' = 2x - 98.

2. Set y' equal to zero and solve for x:
   2x - 98 = 0
   x = 49

3. Compute the second derivative of the function:
   y'' = 2

4. Determine the nature of the critical point x = 49:
   Since the second derivative y'' is positive, the critical point x = 49 corresponds to a local minimum.

Therefore, the only local minimum point for the function y = x^2 - 98x + 4 is (49, -2399).