EVALUATING PIECEWISE FUNCTIONS

Problem 1 :

Evaluate the following.

(a)  18    (b)  17     (c)  14     (d)  20

Solution :

To evaluate f(3), let us observe the functions.

The function will work only if the value of x is lesser than 2.

The function will work only if the value of x is greater than or equal to 2.

f(x) = 5x + 2

f(3) = 5(3) + 2

f(3) = 15 + 2

f(3) = 17

Problem 2 :

Evaluate for the given value.

(a)  -8    (b)  undefined     (c)  8     (d)  -4

Solution :

To evaluate h(-6), let us observe the functions.

The 1st function will work only if the value of x is not equal to -6

The 2nd function will work only if the value of x is equal to -6.

h(x) = x - 2

h(-6) = -6 - 2

h(-6) = -8

Problem 3 :

Evaluate for the given value.

(a)  -2    (b)  17    (c)  4     (d)  -4

Solution :

To evaluate f(-2), let us observe the functions.

When x = -2, which is lesser than -1. So, we choose the second function.

f(x) = -(x - 2)

f(-2) = -(-2-2)

f(-2) = 4

Problem 4 :

Evaluate for the given value.

(i)  f(-4)     (ii)  f(6)      (iii)  f(-2)      (iv)  f(0)

Solution :

 (i)  When x = -4f(x) = 2x + 8f(-4) = 2(-4) + 8f(-4) = -8 + 8f(-4) = 0 (ii)  When x = 6f(x) = √(x + 3)f(6) = √(6 + 3)f(6) = √9f(6)  = ±3 (iii)  When x = -2f(x) = 2x + 8f(-2) = 2(-2) + 8f(-2) = -4 + 8f(-2) = 4 (iv)  When x = 0f(x) = x2 - 3f(0) = 02 - 3f(0) = - 3

Problem 5 :

(i) f(8)      (ii)  f(0)       (iii)  f(4)       (iv)  f(5)

Solution :

 (i)  When x = 8f(x) = |x - 2|f(8) = |8 - 2|f(8) = 6 (ii)  When x = 0f(x) = 2x3 - 1f(0) = 2(0) - 1f(0) = -1 (iii)  When x = 4f(x) = 3 (iv)  When x = 5f(x) = |x - 2|f(5) = |5 - 2|f(5) = 3

Problem 6 :

(i) f(0)      (ii)  f(5)       (iii)  f(2)       (iv)  f(-3)

Solution :

 (i)  When x = 0f(x) = -2x2 - 1f(0) = -2(0)2 - 1f(0) = -1 (ii)  When x = 5f(x) = (4/5)x - 4f(5) = (4/5)5 - 4f(5) = 4 - 4f(5) = 0 (iii)  When x = 2f(x) = -2x2 - 1f(2) = -2(2)2 - 1f(2) = -8 - 1f(2) = -9 (iv)  When x = -3f(x) = -2x2 - 1f(-3) = -2(-3)2 - 1f(-3) = -2(9) - 1f(-3) = -18 - 1f(-3) = -19

Problem 7 :

(i) f(-5)      (ii)  f(11)       (iii)  f(0)       (iv)  f(3)

Solution :

 (i)  When x = -5f(x) = x3 - 7xf(-5) = (-5)3 - 7(-5)= -125 + 35f(-5) = -90 (ii)  When x = 11f(x) = √(2x + 3)f(11) = √(2(11) + 3)f(11) = √(22 + 3)f(11) = √25f(11) = ±5 (iii)  When x = 0f(x) = x3 - 7xf(0) = 03 - 7(0)f(0) = 0 (iv)  When x = 3f(x) = 8f(3) = 8

Recent Articles

1. Finding Range of Values Inequality Problems

May 21, 24 08:51 PM

Finding Range of Values Inequality Problems

2. Solving Two Step Inequality Word Problems

May 21, 24 08:51 AM

Solving Two Step Inequality Word Problems