Answer:
Yes
Step-by-step explanation:
The Mean Value Theorem states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that
[tex]f'(c)=\dfrac{f(b)-f(a)}{b-a}[/tex]
Given [tex]f(x)=x^3+x-9$ in [0,2][/tex]
f(x) is defined, continuous and differentiable.
[tex]f(2)=2^3+2-9=1\\f(0)=0^3+0-9=-9[/tex]
[tex]f'(c)=\dfrac{f(2)-f(0)}{2-0}=\dfrac{1-(-9)}{2}=5[/tex]
[tex]f'(x)=3x^2+1[/tex]
Therefore:
[tex]f'(c)=3c^2+1=5\\3c^2=5-1\\3c^2=4\\c^2=\frac{4}{3} \\c=\sqrt{\frac{4}{3}} =1.15 \in [0,2][/tex]
Since c is in the given interval, the function satisfy the hypotheses of the Mean Value Theorem on the given interval.
