let f(x) = sin(x^3) Find the critical numbers of f on the interval [-2,2]
There are actually a few more critical numbers. We want to find all ways that
`cos(x^3) * 3x^2 = 0`
This is true when x = 0.
We can graph this equation to see all the spots where it is approximately zero.
There are 7 spots where our function equals zero.
So we need to find all ways that `cos(x^3) = 0`
Cos equals zero when at +-pi/2 and +-3pi/2 and +-5pi/2, so we can set x^3 equal to those and find more x values where f'(x) = 0.
`x^3 = pi/2`
`x = (pi/2)^(1/3) ~~ 1.163`
`x^3 = -pi/2`
`x = (-pi/2)^(1/3) ~~ -1.163`
`x^3 = (3pi)/2`
`x = ((3pi)/2)^(1/3) ~~ 1.676`
`x^3 = -(3pi)/2`
`x = ((-3pi)/2)^(1/3) ~~ -1.676`
`x^3 = ((5pi)/2)`
`x = ((5pi)/2)^(1/3) ~~ 1.98`
`x ~~ -1.98`
So those 6 answers are the other spots where f'(x) = 0.
Our set of solutions is x = +- 1.98, +- 1.676, +- 1.163, and 0.
Hope this helps!
A number 'a' is a critical number of f in a given interval if f'(a)=0 or f' is undefined at x=a.
Here, f(x) = sin (x^3)
Differentiating it with respect to x,
f'(x) = 3(x^2)* cos(x^3)
In the interval [-2,2], f' = 0 at x=0.
At all other values of x, cosine will be non-zero and will be defined, so will be the term 3 x^2.
Thus, 0 is the critical number of f in the given interval [-2,2].