Example 1
Write y = x2 + 4x + 1 using function notation and evaluate the function at x = 3.
Solution
Given, y = x2 + 4x + 1
By applying function notation, we get
f(x) = x2 + 4x + 1
Evaluation:
Substitute x with 3
f (3) = 32 + 4 × 3 + 1 = 9 + 12 + 1 = 22
Example 2
Evaluate the function f(x) = 3(2x+1) when x = 4.
Solution
Plug x = 4 in the function f(x).
f (4) = 3[2(4) + 1]
f (4) = 3[8 + 1]
f (4) = 3 x 9
f (4) = 27
Example 3
Write the function y = 2x2 + 4x – 3 in function notation and find f (2a + 3).
Solution
y = 2x2 + 4x – 3 ⟹ f (x) = 2x2 + 4x – 3
Substitute x with (2a + 3).
f (2a + 3) = 2(2a + 3)2 + 4(2a + 3) – 3
= 2(4a2 + 12a + 9) + 8a + 12 – 3
= 8a2 + 24a + 18 + 8a + 12 – 3
= 8a2 + 32a + 27
I think the answer is 1,600 totals students taking metaphysics
CALCULATOR! Sorry but I mean....really?
<span>