intermediate value

The Intermediate Value Theorem says that , in some interval [a, b], if f(a) and f(b) have opposite signs, then f(x) has at least one "root" in this interval. (As long as f(x) is continuous on the interval !!)

So

f(0) = (0)^3 + 4(0) - 4 = -4

and

f(1) = (1)^3 + 4(1) - 4 =  1

Then, at x=0 the function lies below the x axis, and at x =1, the function lies above the x axis........and since polynomials are always continuous, this function must cross the x axis on [0,1]

So...this tells us that this ploynomial has at least one"zero" (root) on the interval [0, 1]....In other words, whatever this value is, it makes f(x) = 0......(the "0" in the problem is correct !!!......)

Thank you Cphill .. 

but I think that this sentence Suffice 

" has at least one solution '' 

right ? 

You want to add in the given domain [0,1] because that is what the question asked for.   

So is my answer correct?  

Yes looks good.

Personally I would say Hence rather than yes but it shouldn't really matter.     

Also, you answer is correct but the original question gave the interval  [0,1]

so personally I would have repeated exactly what they asked for but again it is trivial.