Question

Let $f(x)=\left\{ \begin{array}{ll}1-x^2 & \text{ if }x\leq0 \\ 2x+1 & \text{ if }x\ \textgreater \ 0\end{array}\right.$
Evaluate $f(-2)$ and $f(1)$

Answer 1

Answer:

Step-by-step explanation:

So, remember that f(x) is such that you will have to solve the function based on what is given in the parentheses.  

If it is given as f(-9), x = -9, and likewise for other numbers.

When you get the value for x (example x=9), replace x with 9 wherever it is given in the function.  If it is 1/x, it becomes 1/9.  If it is x squared, it becomes 81 etc.

f(-2) ⇒ x = -2.

∵ x $\leq$ 0, you will solve the f(x) = 1 - $x^{2}$ function.

f (-2) = 1 - (-2)^2

f(-2) = 1 - 4 (Since the minus before the parentheses is still present.)

f(-2) = -3.

f(1) ⇒ x = 1.

∵ x $\geq$ 0, you will solve the f(x) = 2x + 1 function.

f(1) = 2(1) + 1

f(1) = 3.

Answer 2

Step-by-step explanation:

So, remember that f(x) is such that you will have to solve the function based on what is given in the parentheses.

If it is given as f(-9), x = -9, and likewise for other numbers.

When you get the value for x (example x=9), replace x with 9 wherever it is given in the function. If it is 1/x, it becomes 1/9. If it is x squared, it becomes 81 etc.

f(-2) ⇒ x = -2.

∵ x \leq≤ 0, you will solve the f(x) = 1 - x^{2}x

2

function.

f (-2) = 1 - (-2)^2

f(-2) = 1 - 4 (Since the minus before the parentheses is still present.)

f(-2) = -3.

f(1) ⇒ x = 1.

∵ x \geq≥ 0, you will solve the f(x) = 2x + 1 function.

f(1) = 2(1) + 1

f(1) = 3.