StudentAidUnit 4: Functions and Graphs — Exercise 4 2
10th Class Mathematics · Unit 4: Functions and Graphs
4.2.1.Plot graph for the following absolute valued functions: (i) f(x)=|x-2| (ii) f(x)=3|x+3|-4 (iii) f(x)=5|x| (iv) f(x)=|x+2|+3 (v) f(x)=2|x+4|-3 (vi) f(x)=2|x+1|-6
Given
(i)
f(x)=|x-2| = begin{cases} x-2, & xge2 2-x, & x<2 end{cases}
Result
(i) Result
V-shape, vertex (2,0), opens upward
Given
(ii)
f(x)=3|x+3|-4 = begin{cases} 3(x+3)-4, & xge-3 -3(x+3)-4, & x<-3 end{cases}
Result
(ii) Result
V-shape, vertex (-3,-4), opens upward
Given
(iii)
f(x)=5|x| = begin{cases} 5x, & xge0 -5x, & x<0 end{cases}
Result
(iii) Result
V-shape, vertex (0,0), opens upward
Given
(iv)
f(x)=|x+2|+3 = begin{cases} x+2+3, & xge-2 -(x+2)+3, & x<-2 end{cases}
Result
(iv) Result
V-shape, vertex (-2,3), opens upward
Given
(v)
f(x)=2|x+4|-3 = begin{cases} 2(x+4)-3, & xge-4 -2(x+4)-3, & x<-4 end{cases}
Result
(v) Result
V-shape, vertex (-4,-3), opens upward
Given
(vi)
f(x)=2|x+1|-6 = begin{cases} 2(x+1)-6, & xge-1 -2(x+1)-6, & x<-1 end{cases}
Result
(vi) Result
V-shape, vertex (-1,-6), opens upward
4.2.2.Solve and express the solution on number line: (i) |x-2|=6 (ii) |2x+1|=7 (iii) |4x-9|=3 (iv) |7-2x|=1
Formula
(i)
|x-2|=6 implies x-2=6 or x-2=-6
Result
(i) Result
x=8 or x=-4
Formula
(ii)
|2x+1|=7 implies 2x+1=7 or 2x+1=-7
Result
(ii) Result
x=3 or x=-4
Formula
(iii)
|4x-9|=3 implies 4x-9=3 or 4x-9=-3
Result
(iii) Result
x=3 or x=32
Formula
(iv)
|7-2x|=1 implies 7-2x=1 or 7-2x=-1
Result
(iv) Result
x=3 or x=4
4.2.3.Solve and express the solution on number line: (i) |5x+8| \le 3 (ii) |4x-12| \le 0 (iii) |3-4x| > 0 (iv) 1-2|\frac{3}{2}x-5| > -3 (v) |3x-2| \le 12 (vi) |1-2x| > 5
Formula
(i)
|5x+8|le3 implies -3le5x+8le3 implies -11le5xle-5
Result
(i) Result
-115 le x le -1
Formula
(ii)
|4x-12|le0 only when 4x-12=0
Result
(ii) Result
x=3
Formula
(iii)
|3-4x|>0 for all x except 3-4x=0
Result
(iii) Result
x neq 34
(iv)
1-2left|32x-5right|>-3 implies -2left|32x-5right|>-4 implies left|32x-5right|<2
-2<32x-5<2 implies 3<32x<7
(iv) Result
2 < x < 143
Formula
(v)
|3x-2|le12 implies -12le3x-2le12 implies -10le3xle14
Result
(v) Result
-103 le x le 143
Formula
(vi)
|1-2x|>5 implies 1-2x>5 or 1-2x<-5
Result
(vi) Result
x<-2 or x>3