Answer:
The equation does not have a real root in the interval [tex]\rm [0,1][/tex]
Step-by-step explanation:
We can make use of the intermediate value theorem.
The theorem states that if [tex]f[/tex] is a continuous function whose domain is the interval [a, b], then it takes on any value between f(a) and f(b) at some point within the interval. There are two corollaries:
Of course, in our case, we will make use of the first one.
First, we need to proof that our function is continues in [tex]\rm [0,1][/tex], which it is since every polynomial is a continuous function on the entire line of real numbers. Then, we can apply the first corollary to the interval [tex]\rm [0,1][/tex], which means to evaluate the equation in 0 and 1:
[tex]f(x)=x^3-3x+8\\f(0)=8\\f(1)=6[/tex]
Since both values have the same sign, positive in this case, we can say that by virtue of the first corollary of the intermediate value theorem the equation does not have a real root in the interval [tex]\rm [0,1][/tex]. I attached a plot of the equation in the interval [tex]\rm [-2,2][/tex] where you can clearly observe how the graph does not cross the x-axis in the interval.