Question

question
suppose we have a function defined by
f(x)=\begin{cases}x^{2}-6&\text{for }x < 0,\\10 - x&\text{for }xgeq0.end{cases}
what values of (x) give (f(x)=43)?
select the correct answer below:
(x=-7,x = 7.)
(x=-7,x = 7,x=-33.)
(x=-7,x=-33.)
(x=-7)

Explanation:

Step1: Consider the case when $x < 0$

Set $x^{2}-6 = 43$. Then $x^{2}=43 + 6=49$. Taking the square - root of both sides, we get $x=\pm7$. Since $x < 0$, we take $x=-7$.

Step2: Consider the case when $x\geq0$

Set $10 - x=43$. Solving for $x$, we have $x = 10 - 43=-33$. But this value does not satisfy the condition $x\geq0$, so we discard it when considering the $x\geq0$ part of the piece - wise function. So the values of $x$ that give $f(x) = 43$ are $x=-7$ from the $x < 0$ part and $x=-33$ (even though it doesn't fit the $x\geq0$ domain for the second part of the function, when we solve the equation for that part we get this value) and the answer is $x=-7,x = - 33$.

Answer:

C. $x = - 7,x=-33$