Solutions to Problem Set 8

Math 310-01/02

9-22-2017

1. Prove that there is a positive integer n such that $n^2 + n + 3$ is not prime.

There are many possibilities. For example, if $n = 3$ , then

$$n^2 + n + 3 = 15.$$

But 15 is not prime.


2. Suppose that f is a continuous function,

$$f(3) = 2, \quad\hbox{and}\quad f(6) = -3.$$

Prove that $x^2 f(x)$ has a root between $x = 3$ and $x = 6$ .

$g(x) = x^2 f(x)$ is continuous,

$$g(3) = 3^2 f(3) = 18, \quad\hbox{and}\quad g(6) = 6^2 f(6) = -108.$$

By the Intermediate Value Theorem, $g(c) = 0$ for some c between 3 and 6 --- that is, $x^2
   f(x)$ has a root between $x = 3$ and $x = 6$ .


3. Suppose that f is a differentiable function which satisfies

$$f(1) = 3 \quad\hbox{and}\quad f'(x) > 7 \quad\hbox{for all}\quad x.$$

Prove that $f(5) > 31$ .

Apply the Mean Value Theorem to f on the interval $1 \le x \le 5$ . The theorem says that there is a number c such that $1 < c < 5$ and

$$\dfrac{f(5) - f(1)}{5 - 1} = f'(c).$$

Therefore,

$$\eqalign{ \dfrac{f(5) - f(1)}{5 - 1} & = f'(c) > 7 \cr \noalign{\vskip2pt} \dfrac{f(5) - 3}{4} & > 7 \cr \noalign{\vskip2pt} f(5) - 3 & > 28 \cr f(5) & > 31 \cr} \quad\halmos$$


4. Let $f(x) = x^7 + 13 x + 1$ .

(a) Use the Intermediate Value Theorem to prove that $f(x)$ has at least one root.

(b) Use Rolle's theorem to prove that f has only one root.

(a) f is continuous.

$$f(1) = 15 \quad\hbox{and}\quad f(-1) = -13.$$

Since $f(1) > 0$ and $f(-1) < 0$ , the Intermediate Value Theorem implies that f has at least one root between 1 and -1.

(b) Suppose f has two roots a and b with $a < b$ , so $f(a) = f(b)
   = 0$ . By Rolle's Theorem, f has a critical point c between a and b. However,

$$f'(x) = 7 x^6 + 13 > 0 \quad\hbox{for all}\quad x.$$

Hence, f has no critical points. This contradiction shows that f can't have two roots. Since by (a) it has at least one root, it must have only one root.


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