Question

Let f be the function with derivative : sin(x^2+1). How many relative extrema does f have on the interval 2 12 years ago

Answer 1

Set the derivative equal to zero and solve for x:
0 = sin(x² + 1)
kπ = x² + 1
kπ - 1 = x²
±√(kπ - 1) = x

K is an integer, so plug in values for k and solve for x-values in between 2 and 4. When k = 2, 3, 4, and 5 it works out, thus:
x = √(2π - 1), √(3π - 1), √(4π - 1), and √(5π - 1) are critical points

Answer 2

Btw I googled it.

Answer 3

I saw this explanation too but I did not understand the k * pi idea

Answer 4

sin(x² + 1) = 0
x² + 1 = 2πn or x² + 1 = π(2n+1)
x = ± √(2πn - 1) or x = ± √(π(2n+1) - 1)
n = 1 or 2 on the interval (2,4)

Answer: .. 4 relative extrema
This one another answer on google..

Answer 5

I would surely explain you this if I was Maths student ..>.>

Answer 6

well thanks for the help. I can figure it out