Question

For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two decimal places. In each part, enter your answers in increasing order of x values. Round your answers to two decimals.

Answer 1

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Explanation:

We can find the intercept by setting f(x) = 0

`4sin(x^2) = 0`

`sin(x^2) = 0`

`x^2 = n\pi` where n = 0, 1, 2,3, 4, 5,...

`x = \sqrt(n\pi)`

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

`f^(')(x) = 4cos(x^2)2x = 8xcos(x^2) = 0`

`xcos(x^2) = 0`

`x = 0` or

`cos(x^2) = 0`

`x^2 = (\pi)/(2) + n\pi` where n = 0, 1, 2, 3

`x = √(\pi(n+1/2))`

Since we are restricting x between 0 and 3 we can stop at n = 2

So our critical points are at

x = 1.25,
`y = f(1.25) = 4sin(1.25^2) = 4`

x = 2.17 ,
`y = f(2.17) = 4sin(2.17^2) = -4`

x = 2.8,
`y = f(2.8) = 4sin(2.8^2) = 4`

For the inflection point, we can take the 2nd derivative and set it to 0

`f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0`

`cos(x^2) = 2x^2sin(x^2)`

`tan(x^2) = (1)/(2x^2)`

We can solve this numerically to get the inflection points are at

x = 0.81,
`y = f(0.81) = 4sin(0.81^2) = 2.44`

x = 1.81,
`y = f(1.81) = 4sin(1.81^2) = -0.54`

x = 2.52,
`y = f(2.52) = 4sin(2.52^2) = 0.27`