What is the value of the range of the function f(x)=x^2 + 2 for the domain value 1/4?
A. y=2
B. y=2 1/4***
C. y=2 1/8
D. y=2 1/16
Can anyone check my answer?
To determine the value of the function \(f(x)=x^2+2\) for the domain value \(\frac{1}{4}\), we need to substitute \(\frac{1}{4}\) into the function and solve for \(y\). So, let's do that step by step.
Step 1: Substitute \(x=\frac{1}{4}\) into the function:
\[f\left(\frac{1}{4}\right) = \left(\frac{1}{4}\right)^2+2\]
Step 2: Simplify the equation:
\[f\left(\frac{1}{4}\right) = \frac{1}{16}+2\]
\[f\left(\frac{1}{4}\right) = \frac{1}{16}+\frac{32}{16}\]
\[f\left(\frac{1}{4}\right) = \frac{1+32}{16}\]
\[f\left(\frac{1}{4}\right) = \frac{33}{16}\]
Therefore, the value of the function for the domain value \(\frac{1}{4}\) is \(y=\frac{33}{16}\).
Now let's check your answer. The option you selected is B. \(y=2 \frac{1}{4}\). To verify if it matches the calculated value \(\frac{33}{16}\), we can convert the mixed number into an improper fraction:
\[2 \frac{1}{4} = \frac{9}{4}\]
Since \(\frac{33}{16} \neq \frac{9}{4}\), it means that your answer is incorrect.
The correct answer is:
C. \(y=2 \frac{1}{8}\)
I hope this helps!
Nope, however, I'll give you a hint
f(x) = x^2 + 2
Plug in the values...
f(1/4) = (1/4)^2 + 2
Solve the following in order to get your answer.
(1/4)^2 + 2