Question

If the functions
$f(x) \: = 5 {x}^{2} \: +x - 3$
and
$g(x) \: = \frac{2x - 1 }{3}$
find the following functions values.

$1. \: f(0)$
$2. \: f( - 2)$
$3. \: f( \frac{1}{2} )$
$4. \: g(5)$
$5. \: g( - 7)$
$6. \: (0)$




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Answer 1

Answer :

1). f(0) = -3

2). f(-2) = 15

3). f(½) = -5/4

4). g(5) = 3

5). g(-7) = -5

6). g(0) = -⅓

Solution :

→ Given :

• f(x) = 5x² + x - 3

• g(x) = (2x - 1)/3

→ To find :

• f(0)

• f(-2)

• f(½)

• g(5)

• g(-7)

• g(0)

Note :

If the expression of f(x) is given , then to find the value of f(a) simply replace x by a everywhere in the expression of f(x) .

Thus ,

• f(x) = 5x² + x - 3

→ f(0) = 5•0² + 0 - 3

→ f(0) = 0 + 0 - 3

→ f(0) = -3

• f(x) = 5x² + x - 3

→ f(-2) = 5•(-2)² + (-2) - 3

→ f(-2) = 5•4 - 2 - 3

→ f(-2) = 20 - 5

→ f(-2) = 15

• f(x) = 5x² + x - 3

→ f(½) = 5•(½)² + ½ - 3

→ f(½) = 5/4 + 1/2 - 3

→ f(½) = (5 + 2 - 12)/4

→ f(½) = -5/4

• g(x) = (2x - 1)/3

→ g(5) = (2•5 - 1)/3

→ g(5) = (10 - 1)/3

→ g(5) = 9/3

→ g(5) = 3

• g(x) = (2x - 1)/3

→ g(-7) = {2•(-7) - 1}/3

→ g(-7) = (-14 - 1)/3

→ g(-7) = -15/3

→ g(-7) = -5

• g(x) = (2x - 1)/3

→ g(0) = (2•0 - 1)/3

→ g(0) = (0 - 1)/3

→ g(0) = -⅓