Demonstrates that f ( x ) = 4 x⁵ + x³ + 7 x - 2 has accurately one real root.

Solution

From basic Algebra principles we know that since f (x) is a 5^th degree polynomial it will have five roots. What we have to prove here is that only one of those 5 is a real number & the other 4 has to be complex roots.

Firstly, we have to show that it does have at least one real root. To do this note that f (0) = -2 and that f (1) = 10 and thus we can see that f (0) <0 < f (1).  Now, because f (x) is a polynomial we know that this is continuous everywhere and therefore by the Intermediate Value Theorem there is a number c such that 0 < c < 1 and f (c ) < 0 .  In other terms f (x ) has at least one real root.

Now we need to show that this is actually the only real root. To do this we'll utilizes an argument which is called contradiction proof.  What we'll do is suppose that f (x) has two real roots at least.

It means that we can determine real numbers a and b (there might be more, however all we required for this particular argument is two) such that f ( a ) =f (b ) = 0 .  However if we do this then we know from Rolle's Theorem that there has to then be another number c such that

f ′ (c ) = 0 .

However it is a problem. The derivative of this function is,

f ′ ( x ) = 20x⁴ +3x² + 7

Since even the exponents of the first two terms are we know that the first two terms will be greater than or equal to zero always and then we are going to add a positive number onto that and thus we can see that the smallest the derivative will ever be is 7 and this contradicts the statement above that says we ought to have a number c such that f ′ (c) = 0.

We attained these contradictory statements by supposing that f (x) has two roots at least.  Since this supposition leads to a contradiction the supposition has to be false and thus we can only have a single real root.

The cause for covering Rolle's Theorem is that it is required in the proof of the Mean Value Theorem. Following is the theorem.