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Difference between revisions of "1982 AHSME Problems/Problem 19"

(Created page with "== Problem == Let <math>f(x)=|x-2|+|x-4|-|2x-6|</math> for <math>2 \leq x\leq 8</math>. The sum of the largest and smallest values of <math>f(x)</math> is <math>\textbf {(A)...")
 
m (→‎Solution)
 
(6 intermediate revisions by the same user not shown)
Line 17: Line 17:
 
(x-2)+(x-4)-(2x-6) & \mathrm{if} \ 4\leq x\leq8
 
(x-2)+(x-4)-(2x-6) & \mathrm{if} \ 4\leq x\leq8
 
\end{cases},</cmath>
 
\end{cases},</cmath>
−
which simplify to
+
which simplifies to
 
<cmath>f(x) = \begin{cases}
 
<cmath>f(x) = \begin{cases}
 
2x-4 & \mathrm{if} \ 2\leq x<3 \\
 
2x-4 & \mathrm{if} \ 2\leq x<3 \\
Line 24: Line 24:
 
\end{cases}.</cmath>
 
\end{cases}.</cmath>
 
The graph of <math>y=f(x)</math> is shown below.
 
The graph of <math>y=f(x)</math> is shown below.
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(200);
  
−
<b>DIAGRAM NEEDED</b>
+
int xMin = -2;
 +
int xMax = 10;
 +
int yMin = -2;
 +
int yMax = 4;
  
 +
//Draws the horizontal gridlines
 +
void horizontalLines()
 +
{
 +
  for (int i = yMin+1; i < yMax; ++i)
 +
  {
 +
    draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
 +
  }
 +
}
 +
 +
//Draws the vertical gridlines
 +
void verticalLines()
 +
{
 +
  for (int i = xMin+1; i < xMax; ++i)
 +
  {
 +
    draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
 +
  }
 +
}
 +
 +
//Draws the horizontal ticks
 +
void horizontalTicks()
 +
{
 +
  for (int i = yMin+1; i < yMax; ++i)
 +
  {
 +
    draw((-3/16,i)--(3/16,i), black+linewidth(1));
 +
  }
 +
}
 +
 +
//Draws the vertical ticks
 +
void verticalTicks()
 +
{
 +
  for (int i = xMin+1; i < xMax; ++i)
 +
  {
 +
    draw((i,-3/16)--(i,3/16), black+linewidth(1));
 +
  }
 +
}
 +
 +
horizontalLines();
 +
verticalLines();
 +
horizontalTicks();
 +
verticalTicks();
 +
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
 +
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
 +
label("$x$",(xMax,0),(2,0));
 +
label("$y$",(0,yMax),(0,2));
 +
 +
pair A[];
 +
A[0] = (2,0);
 +
A[1] = (3,2);
 +
A[2] = (4,0);
 +
A[3] = (8,0);
 +
 +
draw(A[0]--A[1]--A[2]--A[3],red+linewidth(1.5));
 +
 +
for(int i = 0; i <= 3; ++i) {
 +
dot(A[i],red+linewidth(4.5));
 +
}
 +
 +
label("$(2,0)$",A[0],(0,-1.5),UnFill);
 +
label("$(3,2)$",A[1],(0,1.5),UnFill);
 +
label("$(4,0)$",A[2],(0,-1.5),UnFill);
 +
label("$(8,0)$",A[3],(0,-1.5),UnFill);
 +
</asy>
 
The largest value of <math>f(x)</math> is <math>2,</math> and the smallest value of <math>f(x)</math> is <math>0.</math> So, their sum is <math>\boxed{\textbf {(B)}\ 2}.</math>
 
The largest value of <math>f(x)</math> is <math>2,</math> and the smallest value of <math>f(x)</math> is <math>0.</math> So, their sum is <math>\boxed{\textbf {(B)}\ 2}.</math>
  

Latest revision as of 11:26, 13 September 2021

Problem

Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \leq x\leq 8$. The sum of the largest and smallest values of $f(x)$ is

$\textbf {(A)}\ 1 \qquad \textbf {(B)}\ 2 \qquad \textbf {(C)}\ 4 \qquad \textbf {(D)}\ 6 \qquad \textbf {(E)}\ \text{none of these}$

Solution

Note that at $x=2,3,4,$ one of the three absolute values is equal to $0.$

Without using absolute values, we rewrite $f(x)$ as a piecewise function: \[f(x) = \begin{cases} (x-2)+(4-x)-(6-2x) & \mathrm{if} \ 2\leq x<3 \\ (x-2)+(4-x)-(2x-6) & \mathrm{if} \ 3\leq x<4 \\ (x-2)+(x-4)-(2x-6) & \mathrm{if} \ 4\leq x\leq8 \end{cases},\] which simplifies to \[f(x) = \begin{cases} 2x-4 & \mathrm{if} \ 2\leq x<3 \\ -2x+8 & \mathrm{if} \ 3\leq x<4 \\ 0 & \mathrm{if} \ 4\leq x\leq8 \end{cases}.\] The graph of $y=f(x)$ is shown below. [asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 10; int yMin = -2; int yMax = 4; //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-3/16,i)--(3/16,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-3/16)--(i,3/16), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[]; A[0] = (2,0); A[1] = (3,2); A[2] = (4,0); A[3] = (8,0); draw(A[0]--A[1]--A[2]--A[3],red+linewidth(1.5)); for(int i = 0; i <= 3; ++i) { dot(A[i],red+linewidth(4.5)); } label("$(2,0)$",A[0],(0,-1.5),UnFill); label("$(3,2)$",A[1],(0,1.5),UnFill); label("$(4,0)$",A[2],(0,-1.5),UnFill); label("$(8,0)$",A[3],(0,-1.5),UnFill); [/asy] The largest value of $f(x)$ is $2,$ and the smallest value of $f(x)$ is $0.$ So, their sum is $\boxed{\textbf {(B)}\ 2}.$

~MRENTHUSIASM

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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