Randolph College
Department of Mathematics and Computer Science

 

Homework # 14 - Math 149


The problems on this homework must be done alone.  The honor code is in effect.

 First name:   Last name:


 

1.  Consider the function \(f(x) = x^2\), on the interval \( (-1, 2)\).

a.  This function has no maximum and no minimum on this interval.

b.  This function has a maximum but no minimum on this interval.

c.  This function has a minimum but no maximum on this interval.

d.  This function has a minimum and a maximum on this interval.

 

2.  Consider the function \(f(x) = x^2\), on the interval \( [-2, 1)\).

a.  This function has no maximum and no minimum on this interval.

b.  This function has a maximum but no minimum on this interval.

c.  This function has a minimum but no maximum on this interval.

d.  This function has a minimum and a maximum on this interval.

 

3.  Find all the critical values of the function \(g(x) = x^3-3x^2-9x\).

a.  \(x = 0\) only.

b.  \(x = -1\) only.

c.  \(x = 0\) and \(x = 3\).

d.  \(x = -1\) and \(x = 3\).

e.  None of the above.

 

4.   Can you use the Intermediate Value Theorem to prove that the function \(f(x) = \frac{1}{x-2}\) has a zero (some x value where the function equals zero) somewhere between x = 1 and x = 3?

a.  Yes.

b.  No, because the function \(f\) is not continuous on the interval \( [1, 3] \).

c.  No, because 0 isn't between the values \( f(1)\) and \( f(3)\).

 

5.   Can you use the Intermediate Value Theorem to prove that the function \(f(x) = \frac{1}{x-2}\) has a zero (some x value where the function equals zero) somewhere between x = -1 and x = 1?

a.  Yes.

b.  No, because the function \(f\) is not continuous on the interval \( [-1, 1] \).

c.  No, because 0 isn't between the values \( f(-1)\) and \( f(1)\).

 

6.   Can you use the Intermediate Value Theorem to prove that the function \(f(x) = \frac{x^3+x+1}{x-2}\) has a zero (some x value where the function equals zero) somewhere between x = -1 and x = 1?

a.  Yes.

b.  No, because the function \(f\) is not continuous on the interval \( [-1, 1] \).

c.  No, because 0 isn't between the values \( f(-1)\) and \( f(1)\).