The difference (-) between 3 times a number (3n) and 21.
3n - 21
f(0) = -3(0) + 5 = 5
f(-4) = -3(-4) + 5 = 17
f(6) = ½(6) - 4 = -1
f(9) = 2
f(2) = ½(2) - 4 = -3
Using the restraints, find which equation the f(x) function validates and plug it in.
To find 15% of 130=
130 x 0.15=
19.5
<h2>
Answer:</h2>
The derivative of the function f(x) is:

<h2>
Step-by-step explanation:</h2>
We are given a function f(x) as:

We have:

( Since,
)
Hence, we get:

Also, by using the definition of f'(x) i.e.

Hence, on putting the value in the formula:

Hence, the derivative of the function f(x) is:
