How many points of inflection does the graph of y = sin(x^2) have on interval [-pi, pi]?

Question

I know the answer is 8, but how do you find the zeros of the second derivative?

Answer 1

The positive x co-ordinates of the points of inflection are (approximately) ,808, 1.814, 2.522 and 3.079; these were obtained by solving f"(x)=0 using interval halving on a spreadsheet. The negative values are symmetric about the y axis. These values look correct using a pair of graphs on DESMOS.

I do not agree with Stanton D.'s comment that the values can be obtained from the points of inflection of sin x. For one thing, x=0 is not a point of inflection of f(x) and for another, the points I obtained are not equal to the square root of the values Stanton D. proposed. I cannot explain why Stanton D's comment does not work, but it appears not to do so.

Answer 2

Hi Asked,

there are only 7, not 8, points. When x^2 has values 0, pi, 2pi, and 3pi, the sin curve has inflections (and zero crossings!). The value of (pi)^2 at 9.87 is just a little beyond 3pi (9.42), so that's all of them. These x^2 values result from x values of -(3pi)^0.5, -(2pi)^0.5 , -(pi)^0.5, 0, pi^0.5, (2pi)^0.5, and (3pi)^0.5 . Really, pulling out the chain rule and starting to crank is overkill!

-- Cheers, --Mr. d.

Answer 3

y^' = 2x·cos(x²)

y^" = 2cos(x²) - 4x²sin(x²) = 0

cos(x²) - 2x²sin(x²) = 0

But solving this analytically I am unsure about. It is an even function, so the 4 positive roots between 0 and π are repeated in the negative interval. Difficult to imagine an easy algebraic solution, given that we have powers of x and trig functions of x in the same equation.

Dividing by cos(x²) gives 1 - 2x²tan(x²) = 0 and x²tan(x²) = 1/2 and with z = x² we get ztan(z) = 1/2 but after?