Sure! Let's carefully analyze the function [tex]\( f(t) = \frac{t^2 + 3t - 10}{t - 2} \)[/tex] and find the value of [tex]\( f(2) \)[/tex] that would make this function continuous at [tex]\( t = 2 \)[/tex].
### Step-by-Step Solution:
1. Identify the Point of Interest:
We are interested in the behavior of the function [tex]\( f(t) \)[/tex] as [tex]\( t \)[/tex] approaches 2.
2. Simplify the Expression:
Simplify the function [tex]\( f(t) \)[/tex] wherever possible. Notice that the numerator [tex]\( t^2 + 3t - 10 \)[/tex] can be factored:
[tex]\[
t^2 + 3t - 10 = (t + 5)(t - 2)
\][/tex]
Therefore, the function can be rewritten as:
[tex]\[
f(t) = \frac{(t + 5)(t - 2)}{t - 2}
\][/tex]
3. Simplify the Function:
As long as [tex]\( t \neq 2 \)[/tex], we can cancel out the common factor of [tex]\( t - 2 \)[/tex]:
[tex]\[
f(t) = t + 5 \quad \text{for} \quad t \neq 2
\][/tex]
4. Find the Limit as [tex]\( t \)[/tex] Approaches 2:
To determine the appropriate value for [tex]\( f(2) \)[/tex] that will make [tex]\( f(t) \)[/tex] continuous at [tex]\( t = 2 \)[/tex], we need to find the limit of [tex]\( f(t) \)[/tex] as [tex]\( t \)[/tex] approaches 2:
[tex]\[
\lim_{t \to 2} f(t) = \lim_{t \to 2} (t + 5)
\][/tex]
Since [tex]\( t + 5 \)[/tex] is a continuous function, we can simply substitute [tex]\( t = 2 \)[/tex] into the expression:
[tex]\[
\lim_{t \to 2} (t + 5) = 2 + 5 = 7
\][/tex]
5. Define [tex]\( f(2) \)[/tex] for Continuity:
For [tex]\( f(t) \)[/tex] to be continuous at [tex]\( t = 2 \)[/tex], the value of the function at [tex]\( t = 2 \)[/tex] must equal the limit found above:
[tex]\[
f(2) = 7
\][/tex]
### Conclusion:
By defining [tex]\( f(2) \)[/tex] as 7, we ensure that the function [tex]\( f(t) \)[/tex] is continuous at [tex]\( t = 2 \)[/tex]. So, [tex]\( f(2) = 7 \)[/tex] is the value that extends [tex]\( f(t) = \frac{t^2 + 3t - 10}{t - 2} \)[/tex] to be continuous at [tex]\( t = 2 \)[/tex].
