Question

suppose that ( f(t) = t^2 + 2t - 3 ). (a) what is the average rate of change of ( f(t) ) over the interval 1 to 2? (b) what is the (instantaneous) rate of change of ( f(t) ) when ( t = 1 )? the average rate of change of ( f(t) ) over the interval 1 to 2 is 5. (simplify your answer.) the (instantaneous) rate of change of ( f(t) ) when ( t = 1 ) is (square). (simplify your answer.)

Explanation:

Part (b)

Step1: Find the derivative of \( f(t) \)

The function is \( f(t) = t^2 + 2t - 3 \). Using the power rule for differentiation, if \( y = x^n \), then \( y' = nx^{n - 1} \). For the first term \( t^2 \), the derivative is \( 2t \); for the second term \( 2t \), the derivative is \( 2 \); and the derivative of the constant term \( -3 \) is \( 0 \). So, \( f'(t) = 2t + 2 \).

Step2: Evaluate the derivative at \( t = 1 \)

Substitute \( t = 1 \) into \( f'(t) \): \( f'(1) = 2(1) + 2 \).

Step3: Simplify the expression

\( 2(1) + 2 = 2 + 2 = 4 \).

Answer:

The (instantaneous) rate of change of \( f(t) \) when \( t = 1 \) is \( \boldsymbol{4} \).