given $f(x)=\begin{cases}3sqrt{2x - 1}&\text{for }xleq2\\6(2x - 1)^2-3&\text{for }x > 2end{cases}$. what is…
given $f(x)=\begin{cases}3sqrt{2x - 1}&\text{for }xleq2\\6(2x - 1)^2-3&\text{for }x > 2end{cases}$. what is $limlimits_{x\rightarrow2^{+}}f(x)$?\n$3sqrt{3}$\n$3sqrt{5}$\n$36$\n$51$
Answer
Explanation:
Step1: Identify the correct - part of the function
Since we are finding $\lim_{x\rightarrow2^{+}}f(x)$, we use the part of the function for $x > 2$, which is $f(x)=6(2x - 1)^{2}-3$.
Step2: Substitute $x = 2$ into the function
Substitute $x = 2$ into $y=6(2x - 1)^{2}-3$. First, calculate the value inside the parentheses: $2x-1$ when $x = 2$ is $2\times2-1=3$. Then, square it: $3^{2}=9$. Multiply by 6: $6\times9 = 54$. Finally, subtract 3: $54-3=51$.
Answer:
51