#1**+10 **

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 !!!......)

CPhill
Oct 10, 2014

#1**+10 **

Best Answer

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 !!!......)

CPhill
Oct 10, 2014

#2**0 **

Thank you Cphill ..

but I think that this sentence Suffice

" has at least one solution ''

xvxvxv
Oct 10, 2014