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For the function f(t) = t(t-3)/(t-2) + (4-t)/(t-2), find the values of f(0), f(-3), and f(2).

2 Answers

3 votes

Answer:

f(0)=-2, f(-3)=-5, f(2)=undefined

Explanation:

just replace t with whatever number

User Hristo Eftimov
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1 vote

f(0) = -2, f(-3) = -5, f(2) = 2 (using limit) as the function is undefined at 2.

Here are the values of f(0), f(-3), and f(2) for the given function:

f(0) = -2

Substitute t = 0 into the function:

f(0) = (0)(0-3)/(0-2) + (4-0)/(0-2) = -3/-2 + 4/-2 = -2

f(-3) = -5

Substitute t = -3 into the function:

f(-3) = (-3)(-3-3)/(-3-2) + (4-(-3))/(-3-2) = 18/-5 + 7/-5 = -5

f(2) = 2 (using the limit as t approaches 2)

The function is undefined at t = 2 due to division by zero. However, we can find the limit as t approaches 2 to determine the value:

lim(t→2) f(t) = lim(t→2) [t(t-3)/(t-2) + (4-t)/(t-2)]

This limit evaluates to 2 using L'Hôpital's rule or by simplifying the expression.

Therefore, f(0) = -2, f(-3) = -5, and f(2) = 2 (using the limit).

User BCDeWitt
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