Answer
The values $d=±\sqrt 5$ are not in the interval $(−2,2)$, this contradicts the assumption that there are two distinct roots in the interval $[−2,2]$.
Hence, $f(x)=0$ has at most one solution in $[−2,2]$
Work Step by Step
Suppose that $f(x)= x^{3}−15x+c=0$ has 2 solutions $x=a, x=b, f(a)=f(b)=0$
By the Rolle Theorem, there exists a number $d∈(a,b)$ such that $f′(d)=0=3d^{2}−15$
So, $d=±\sqrt 5$
But the values are not in the interval $(−2,2)$, this contradicts the assumption that there are two distinct roots in the interval $[−2,2]$.
Hence, $f(x)=0$ has at most one solution in $[−2,2]$
