Factor this problem and then use the Zero Product property to get the two solutions 6x²-23x+20
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
The <span>given the piecewise function is :
</span>
![f(x) = \[ \begin{cases} 2x & x \ \textless \ 1 \\ 5 & x=1 \\ x^2 & x\ \textgreater \ 1 \end{cases} \]](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5C%5B%20%5Cbegin%7Bcases%7D%20%0A%20%20%20%20%20%202x%20%26%20x%20%5C%20%5Ctextless%20%5C%20%201%20%5C%5C%0A%20%20%20%20%20%205%20%26%20x%3D1%20%5C%5C%0A%20%20%20%20%20%20x%5E2%20%26%20x%5C%20%5Ctextgreater%20%5C%201%20%0A%20%20%20%5Cend%7Bcases%7D%0A%5C%5D)
To find f(5) ⇒ substitute with x = 5 in the function → x²
∴ f(5) = 5² = 25
To find f(2) ⇒ substitute with x = 5 in the function → x²
∴ f(2) = 2² = 4
To find f(-2) ⇒ substitute with x = 5 in the function → 2x
∴ f(-2) = 2 * (-2) = -4
To find f(1) ⇒ substitute with x = 1 in the function → 5
∴ f(1) = 5
================================
So, the statements which are true:<span>
</span><span></span>
</span>
To find f(5) ⇒ substitute with x = 5 in the function → x²
∴ f(5) = 5² = 25
To find f(2) ⇒ substitute with x = 5 in the function → x²
∴ f(2) = 2² = 4
To find f(-2) ⇒ substitute with x = 5 in the function → 2x
∴ f(-2) = 2 * (-2) = -4
To find f(1) ⇒ substitute with x = 1 in the function → 5
∴ f(1) = 5
================================
So, the statements which are true:<span>