Question

How do you determine all values of c that satisfy the mean value theorem on the interval [-1,1] for `f(x) = 3x^5+5x^3+15x`?

Answer 1

There are two values `c=+-0.62`

Explanation:

On the interval [-1,1], the function `f(x)` is defined and continuous, and differentiable on the interval [-1,1] as it is a polynomial function. So we can apply the mean value theorem.

There is value `c ∈ [-1,1]` such that
`f'(c)=(f(1)-f(-1))/(1-(-1))`

Let's determine `f(1)=3+5+15=23`
And `f(-1)=-3-5-15=-23`

So `f'(c)=(23--23)/2=46/2=23`

Let's calculate `f'(x)=15x^4+15x^2+15=15(x^4+x^2+1)`
Then `f'(c)=15(c^4+c^2+1)=23`

`c^4+c^2+1=23/15`
`c^4+c^2-8/15=0`
Rewrite as `15c^4+15c^2-8=0`
let `c^2=y`
`15y^2+15y-8=0`
`Delta = 15^2+4*15*8=705`
So `y=(-15+-sqrt705)/30=(-15+26.6)/30=11.6/30=0.39`

We take only the positive root

Then `c=+-sqrt0.39=+-0.62`

`+-0.62 ∈[-1,1]`