Answer:
[tex]f'(x)=3-9x^{2}[/tex] and [tex]f''(x)=-18x[/tex]
Step-by-step explanation:
In order to find the derivatives, first we need to remember that for polynomial functions:
[tex]f'(x)=(x^{n}+x^{m})'= (x^{n})'+(x^{m})'[/tex], as well as that:
[tex]f'(x)= (x^{n})' = n*(x^{n-1})[/tex]
1. First derivative of the function:
[tex]f(x)=9+3x-3x^{3}[/tex]
[tex]f'(x)=(9)'+(3x)'-(3x^{3})'[/tex] using the property [tex]f'(x)= (x^{n})' = n*(x^{n-1})[/tex] then
[tex]f'(x)=3-3*3x^{2}[/tex], remember that the derivative of a constant is equal to 0
[tex]f'(x)=3-9x^{2}[/tex]
2. Second derivative:
[tex]f'(x)=3-9x^{2}[/tex]
[tex]f''(x)=(3-9x^{2})'[/tex] using the property [tex]f'(x)= (x^{n})' = n*(x^{n-1})[/tex] then
[tex]f''(x)=(3)'-(9x^{2})'[/tex]
[tex]f''(x)=-(9*2)x^{1}[/tex]
[tex]f''(x)=-18x[/tex]
In conclusion, [tex]f'(x)=3-9x^{2}[/tex] and [tex]f''(x)=-18x[/tex]
