Question

(a) Suppose that \(f\) is differentiable on \(\mathbb{R}\) and has two roots. Show that \({f^\prime }\) has at least one root.

(b) Suppose is \(f\) twice differentiable on \(\mathbb{R}\) and has three roots. Show that \({f^{\prime \prime }}\) has at least one real root.

(c) Can you generalize parts (a) and (b)?

Short answer

(a) The function \({f^\prime }\) has at least one root if \(f\) is differentiable on \(p\) and has two roots.

(b) The function \({f^{\prime \prime }}\) has at least one root if \(f\) is twice differentiable on \(\mathbb{R}\) and has three roots.

(c) It can be generalized that if \(f\) is \(n\) times differentiable on \(\mathbb{R}\) and has \(n + 1\) roots, then \({f^{(n)}}\) has at least one root.

Step-by-step solution

Given data

The function \(f\) is differentiable and twice differential on \(\mathbb{R}\).

Concept of Rolle’s Theorem

“Let a function\(f\)satisfies the following conditions,

1. A function\(f\)is continuous on the closed interval\((a\;,\;b)\).

2. A function\(f\)is differentiable on the open interval\((a\;,\;b)\).

3.\(f(a) = f(b)\)

Then, there is a number\(c\)in open interval\((a\;,\;b)\)such that\({f^\prime }(c) = 0\).”

Calculation to show the function \({f^\prime }\) has at least one root

(a)

From the given statement, \(f\) is differentiable on \(\mathbb{R}\) which means that \({f^\prime }\)exists and since \(f\) is differentiable, \(f\) is obviously continuous on \(\mathbb{R}\).

Since \(f\) has two roots that is \(f(a) = f(b) = 0\) with \(a < b\), by Rolle's Theorem there exist a number with \(a < c < b\) such that \({f^\prime }(c) = 0\).

Hence, the given statement is showed.

Calculation to show the function \({f^{\prime \prime }}\) has at least one root

(b)

From the given statement \(f\) is twice differentiable on \(\mathbb{R}\) which means that \({f^{\prime \prime }}\) exists and since \(f\) is differentiable, \(f\) is obviously continuous on \(\mathbb{R}\).

Since \(f\) has three roots that is \(f(a) = f(b) = f(c) = 0\) with \(a < b < c\).

So, Rolle's Theorem applied to \(f(x)\) on \((a\,,\;b)\) and \((b\;,\;c)\) there exist a number \(d\) and \(e\) with \(a < d < b\) and \(b < e < c\) respectively such that \({f^\prime }(d) = {f^\prime }(e) = 0.\)

Thus, \({f^{\prime \prime }}\) have two roots.

Hence the given statement is showed.

Calculation to compare part (a) and part (b)

(c)

From part (a) and part (b) observe that:

If \(f\) is \(1\) time differentiable on \(\mathbb{R}\) and has \(2\) roots, then, \({f^\prime }\) has at least one root.

If \(f\) is \(2\) time differentiable on \(\mathbb{R}\) and has \(3\) roots, then, \({f^{\prime \prime }}\) has at least one root.

From these two statements, it is can be generalized that if \(f\) is \(n\) times differentiable on \(\mathbb{R}\) and has \(n + 1\) roots, then \({f^{(n)}}\) has at least one root.