Find f(-5) and f(12) for each piecewise function f(x){-4x+3 if x<3
{-x^3 If 3<_x<_8
{3x^2+1 If x>8
The lines go under the inequality lol but please help !

Answers

Answer 1
Answers:f(-5) = 23f(12) = 433

==================================================

Explanation:

The given piecewise function is

[tex]f(x) = \begin{cases}-4x+3 \ \text{ if } \ x < 3\\-x^3 \ \text{ if } \ 3 \le x \le 8\\3x^2+1 \ \text{ if } \ x > 8\end{cases}[/tex]

At first piecewise functions may be strange confusing things, but they aren't so bad. I like to think of it like this: f(x) is a function that changes its identity based on what the input x is. We have three situations

f(x) = -4x+3 when x < 3f(x) = -x^3 when [tex]3 \le x \le 8[/tex]f(x) = 3x^2+1 when x > 8

In a sense, we have three different functions but they are combined somehow.

If x is smaller than 3, then we go for the first definition. Or if x is between 3 and 8, then we go for the second definition. Or if x is larger than 8, then we go for the third definition.

-----------------------

f(-5) means f(x) when x = -5. We see that -5 is smaller than 3, so x = -5 makes x < 3 true. We'll use the first definition

f(x) = -4x+3

f(-5) = -4(-5)+3

f(-5) = 20+3

f(-5) = 23

------------------------

Now the input is x = 12. This is larger than 8. In other words, x = 12 makes x > 8 true. We'll use the third definition

f(x) = 3x^2+1

f(12) = 3(12)^2+1

f(12) = 3(144)+1

f(12) = 432+1

f(12) = 433

------------------------

Side notes:

We won't use the second definition since we don't have any x inputs between 3 and 8To say "less than or equal to" on a keyboard, you can write "<=" without quotes. For example, [tex]x \le 5[/tex] is the same as x<=5

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Hey there!!

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