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Section 4-7 : The Mean Value Theorem

6. Show that \(f\left( x \right) = {x^3} - 7{x^2} + 25x + 8\) has exactly one real root.

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Hint : Can you use the Intermediate Value Theorem to prove that it has at least one real root?
Start Solution

First let’s note that \(f\left( 0 \right) = 8\). If we could find a function value that was negative the Intermediate Value Theorem (which can be used here because the function is continuous everywhere) would tell us that the function would have to be zero somewhere. In other words, there would have to be at least one real root.

Because the largest power of \(x\) is 3 it looks like if we let \(x\) be large enough and negative the function should also be negative. All we need to do is start plugging in negative \(x\)’s until we find one that works. In fact, we don’t even need to do much : \(f\left( { - 1} \right) = - 25\).

So, we can see that \( - 25 = f\left( { - 1} \right) < 0 < f\left( 0 \right) = 8\) and so by the Intermediate Value Theorem the function must be zero somewhere in the interval \(\left( { - 1,0} \right)\). The interval itself is not important. What is important is that we have at least one real root.

Hint : What would happen if there were more than one real root?
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Next, let’s assume that there is more than one real root. Assuming this means that there must be two numbers, say \(a\) and \(b\), so that,

\[f\left( a \right) = f\left( b \right) = 0\]

Next, because \(f\left( x \right)\) is a polynomial it is continuous and differentiable everywhere and so we could use Rolle’s Theorem to see that there must be a real value, \(c\), so that,

\[f'\left( c \right) = 0\]

Note that Rolle’s Theorem tells us that \(c\) must be between \(a\) and \(b\). Since both of these are real values then \(c\) must also be real.

Hint : Is that possible?
Show Step 3

Because \(f\left( x \right)\) is a polynomial it is easy enough to see if such a \(c\) exists.

\[f'\left( x \right) = 3{x^2} - 14x + 25\hspace{0.5in} \to \hspace{0.5in}3{c^2} - 14c + 25 = 0\hspace{0.5in} \to \hspace{0.5in}c = \frac{{7 \pm \sqrt {26} \,i}}{3}\]

So, we can see that in fact the only two places where the derivative is zero are complex numbers and so are not real numbers. Therefore, it is not possible for there to be more than one real root.

From Step 1 we know that there is at least one real root and we’ve just proven that we can’t have more than one real root. Therefore, there must be exactly one real root.