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 = -4

f(x) = 2x + 8

f(-4) = 2(-4) + 8

f(-4) = -8 + 8

f(-4) = 0

(ii)  When x = 6

f(x) = √(x + 3)

f(6) = √(6 + 3)

f(6) = √9

f(6)  = ±3

(iii)  When x = -2

f(x) = 2x + 8

f(-2) = 2(-2) + 8

f(-2) = -4 + 8

f(-2) = 4

(iv)  When x = 0

f(x) = x2 - 3

f(0) = 02 - 3

f(0) = - 3

Problem 5 :

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

Solution :

(i)  When x = 8

f(x) = |x - 2|

f(8) = |8 - 2|

f(8) = 6

(ii)  When x = 0

f(x) = 2x3 - 1

f(0) = 2(0) - 1

f(0) = -1

(iii)  When x = 4

f(x) = 3

(iv)  When x = 5

f(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 = 0

f(x) = -2x2 - 1

f(0) = -2(0)2 - 1

f(0) = -1

(ii)  When x = 5

f(x) = (4/5)x - 4

f(5) = (4/5)5 - 4

f(5) = 4 - 4

f(5) = 0

(iii)  When x = 2

f(x) = -2x2 - 1

f(2) = -2(2)2 - 1

f(2) = -8 - 1

f(2) = -9

(iv)  When x = -3

f(x) = -2x2 - 1

f(-3) = -2(-3)2 - 1

f(-3) = -2(9) - 1

f(-3) = -18 - 1

f(-3) = -19

Problem 7 :

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

Solution :

(i)  When x = -5

f(x) = x3 - 7x

f(-5) = (-5)3 - 7(-5)

= -125 + 35

f(-5) = -90

(ii)  When x = 11

f(x) = √(2x + 3)

f(11) = √(2(11) + 3)

f(11) = √(22 + 3)

f(11) = √25

f(11) = ±5

(iii)  When x = 0

f(x) = x3 - 7x

f(0) = 03 - 7(0)

f(0) = 0

(iv)  When x = 3

f(x) = 8

f(3) = 8

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