Functions, Equations and Polynomials

Toolkit

Functions

A function is a set of ordered pairs $(x,y)$, in which for every $x$ value, there is exactly one $y$ value. If we graph the function, then it must pass the Vertical Line Test, which says that if we draw a vertical line anywhere on the graph, it will pass through at most one point. We use the notation $f(x)=y$ to state that $y$ is the value of the function that corresponds to $x$.

Composition of functions is the process of combining two or more functions, where one function is performed first and the result is substituted in place of $x$ into the next function and so on. The composition of functions $f$ and $g$ is written as $f(g(x))$. In the notation $f(g(x))$, the function $g$ is applied first, followed by the function $f$.

The inverse of a function undoes the action of the function. That is, if the function is a set of ordered pairs $(x,y)$, then the inverse will be set of ordered pairs $(y,x)$. If the inverse of a function $f(x)$ is also a function, it is denoted $f^{-1}(x)$.

• $f^{-1}(f(x))=x$ for all $x$ values in the domain of $f(x)$

• $f(f^{-1}(x))=x$ for all $x$ values in the domain of $f^{-1}(x)$.

Solving Equations, Inequalities and Systems of Equations

You should be able to solve an equation, an inequality or a system of equations algebraically. When solving a system of equations, the method of elimination or the method of substitution are two methods that can be useful.

Parabolas

The quadratic polynomial $f(x)=a{x}^2+bx+c$ (with $a$,$b$,$c$ real and $a\ne 0$) has two roots given by the quadratic formula, $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$. These roots are

• real and distinct, if $b^2-4ac>0$

• real and equal, if $b^2-4ac=0$

• distinct and non-real, if $b^2-4ac<0$

The sum of these two roots is $-{\displaystyle \frac{b}{a}}$ and their product is ${\displaystyle \frac{c}{a}}$.

Since $y=a{x}^2+bx+c=a{(x+{\displaystyle \frac{b}{2a}})}^2+{\displaystyle \frac{4ac-b^2}{4a}}$, the parabola’s vertex is located at $(-{\displaystyle \frac{b}{2a}},{\displaystyle \frac{4ac-b^2}{4a}})$.

Polynomials

• The Remainder Theorem states that when a polynomial $p(x)=a_0+a_1{x}^1+\cdots +a_n{x}^n$, of degree $n$, is divided by $(x-k)$ the remainder is $p(k)$.

• The Factor Theorem, which follows from the Remainder Theorem, states that $p(k)=0$ if and only if $(x-k)$ is a factor of $p(x)$.

• A polynomial equation $p(x)=0$, where $p(x)$ has degree $n$, has at most $n$ real roots.

• The Rational Roots Theorem states that if we have a polynomial $p(x)$ with integer coefficients, then the rational roots of the polynomial are of the form ${\displaystyle \frac{q}{r}}$, where $q$ is a factor of the constant term of $p(x)$ and $r$ is a factor of the leading coefficient of $p(x)$ (the coefficient of the term of the highest degree).

Sample Problems

1. For each positive real number $x$, define $f(x)$ to be the number of prime numbers $p$ that satisfy $x\le p\le x+10$. What is the value of $f(f(20))$?

   Solution

   Let $a=f(20)$. Then $f(f(20))=f(a)$. To calculate $f(f(20))$, we determine the value of $a$ and then the value of $f(a)$. By definition, $a=f(20)$ is the number of prime numbers $p$ that satisfy $20\le p\le 30$. The prime numbers between $20$ and $30$, inclusive, are $23$ and $29$, so $a=f(20)=2$. Thus, $f(f(20))=f(a)=f(2)$. By definition, $f(2)$ is the number of prime numbers $p$ that satisfy $2\le p\le 12$. The prime numbers between $2$ and $12$, inclusive, are $2$, $3$, $5$, $7$, $11$, of which there are $5$. Therefore, $f(f(20))=5.$

2. If $x^2-x-2=0$, determine all possible values of $1-{\displaystyle \frac{1}{x}}-{\displaystyle \frac{6}{x^2}}$.

   Solution

   We have $$1-{\displaystyle \frac{1}{x}}-{\displaystyle \frac{6}{x^2}}={\displaystyle \frac{x^2-x-6}{x^2}}={\displaystyle \frac{x^2-x-2-4}{x^2}}$$ Since $x^2-x-2=0$, we have $${\displaystyle \frac{x^2-x-2-4}{x^2}}=-{\displaystyle \frac{4}{x^2}}$$ Factoring, we obtain $x^2-x-2=(x-2)(x+1)=0$. Thus, $x=2$ or $x=-1$. Therefore, the possible values of the expression are $-1$ and $-4$.

3. Given that $x^2=8x+y$ and $y^2=x+8y$ with $x\ne y$, determine the value of $x^2+y^2$.

   Solution

   Adding the two equations we obtain $x^2+y^2=9x+9y$. Subtracting the second equation from the first we obtain $x^2-y^2=7x-7y$. Factoring both sides we obtain $(x-y)(x+y)=7(x-y)$. Since $x\ne y$, we have that $x-y\ne 0$ and so we can divide both sides by $x-y$ to obtain that $x+y=7$. Therefore, $x^2+y^2=9(x+y)=9(7)=63$.

4. If the graph of the parabola $y=x^2$ is translated to a position such that its $x$ intercepts are $-d$ and $e$ and its $y$ intercept is $-f$, (where $d$,$e$,$f>0)$, show that $de=f$.

   Solution 1 (easy)

   Since the $x$ intercepts are $-d$ and $e$, the parabola must be of the form $y=a(x+d)(x-e)$. Also, since we have only translated $y=x^2$, it follows that $a=1$. When $x=0$, we obtain the $y$-intercept. Therefore, setting $x=0$ gives $-f=-de$ and the result follows.

   Solution 2 (harder)

   Let the parabola be $y=a{x}^2+bx+c$. Now, as in the first solution, $a=1$. Then solving for the $x$- and $y$-intercepts we find $e={\displaystyle \frac{-b+\sqrt{b^2-4c}}{2}}$, $-d={\displaystyle \frac{-b-\sqrt{b^2-4c}}{2}}$ and$-f=c$. Now multiplication of these two expressions gives $-de={\displaystyle \frac{-b-\sqrt{b^2-4c}}{2}}\cdot {\displaystyle \frac{-b+\sqrt{b^2-4c}}{2}}={\displaystyle \frac{b^2-b^2+4c}{4}}=c=-f$ as required.

5. Find all values of $x$ such that $x+{\displaystyle \frac{36}{x}}\ge 13$.

   Solution

   First, we note that $x\ne 0$. If $x>0$, we can multiply the inequality by this positive quantity and arrive at $x^2-13x+36\ge 0$ or $(x-4)(x-9)\ge 0$. We have two cases to consider. The first has $x-4\ge 0$ and $x-9\ge 0$ and the second has $x-4\le 0$ and $x-9\le 0$.

   In the first case, the two inequalities combine to give $x\ge 9$. In the second case, the two inequalities combine to give $x\le 4$. We also have that $x>0$, and so this gives $0<x\le 4$ or $x\ge 9$.

   If $x<0$, the left side of the inequality is negative, which means it is not greater than $13$. Therefore, $0<x\le 4$ or $x\ge 9$.

6. If a polynomial leaves a remainder of 5 when divided by $x-3$ and a remainder of $-7$ when divided by $x+1$, what is the remainder when the polynomial is divided by $x^2-2x-3$?

   Solution

   We observe that when we divide a polynomial by a second degree polynomial the remainder will be a linear polynomial or a constant polynomial. Thus, the division statement becomes $$\begin{array}{}\text{(}\ast \text{)}& p(x)& =(x^2-2x-3)q(x)+ax+b\end{array}$$ where $p(x)$ is the polynomial, $q(x)$ is the quotient polynomial and $ax+b$ is the remainder. Now we observe that the remainder theorem states that $p(3)=5$ and $p(-1)=-7$. Also we notice that $x^2-2x-3=(x-3)(x+1)$. Thus, substituting $x=3$ and $-1$ into $(\ast )$ we obtain $$\begin{array}{rl}p(3)& =5=3a+b\\ p(-1)& =-7=-a+b\end{array}$$ Solving the resulting system of equations gives $a=3$ and $b=-4$. Therefore, the remainder is $3x-4$.

Problem Set

1. If $x$ and $y$ are real numbers, determine all solutions $(x,y)$ to the system of equations $$\begin{array}{rl}{x}^2-xy+8& =0\\ x^2-8x+y& =0\end{array}$$

2. The parabola defined by the equation $y=(x-1{)}^2-4$ intersects the $x$-axis at points $P$ and $Q$. If $(a$,$b)$ is the midpoint of $PQ$, what is the value of $a$?

3. 1. The equation $y=x^2+2ax+a$ represents a parabola for all real values of $a$. Prove that there exists a common point through which all of these parabolas pass, and determine the coordinates of this point.

   2. The vertices of these parabolas lie on a curve. Prove that this curve is itself a parabola whose vertex is the common point found in part (a).

4. Determine all real values of $p$ and $r$ that satisfy the following system of equations. $$\begin{array}{rl}p+pr+p{r}^2& =26\\ p^{2}r+p^2{r}^2+p^2{r}^3& =156\end{array}$$

5. A quadratic equation $a{x}^2+bx+c=0$ (where $a$, $b$, and $c$ are not zero), has real roots. Prove that $a,b,c$, in that order, cannot be consecutive terms in a geometric sequence.

6. A quadratic equation $a{x}^2+bx+c=0$ (where $a$, $b$, and $c$ are integers and $a\ne 0$), has integer roots. If $a,b,c$, in that order, are consecutive terms in an arithmetic sequence, solve for the roots of the equation.

7. Solve the following equation for $x$. $$(x^2-3x+1{)}^2-3(x^2-3x+1)+1=x.$$

8. The parabola $y=(x-2{)}^2-16$ has its vertex at point $A$ and its larger $x$-intercept at point $B$. Find the equation of the line through $A$ and $B$.

9. Solve the equation $(x-b)(x-c)=(a-b)(a-c)$ for $x$.

10. Given that $x=-2$ is a solution of $x^3-7x-6=0$, find the other solutions.

11. Find the value of $a$ such that the equation below in $x$ has real roots, the sum of whose squares is a minimum. $$4x^2+4(a-2)x-8a^2+14a+31=0$$

12. If $f(x)={\displaystyle \frac{3x-7}{x+1}}$ and $g(x)$ is the inverse of $f(x)$, then determine the value of $g(2)$.

13. If $(-2$,$7)$ is the maximum point for the function $y=-2x^2-4ax+k$, determine $k$.

14. The roots of $x^2+cx+d=0$ are $a$ and $b$ and the roots of $x^2+ax+b=0$ are $c$ and $d$. If $a$, $b$, $c$ and $d$ are nonzero, calculate $a+b+c+d$.

15. If $y=x^2-2x-3$, then determine the minimum value of ${\displaystyle \frac{y-4}{(x-4{)}^2}}$.

16. Suppose that the function $g$ satisfies $g(x)=2x-4$ for all real numbers $x$ and that $g^{-1}$ is the inverse function of $g$. Suppose that the function $f$ satisfies $g(f(g^{-1}(x)))=2x^2+16x+26$ for all real numbers $x$. What is the value of $f(\pi )$?