Question

introduction to derivatives: problem 2 (1 point) use the definition of the derivative to find the derivative of: $f(t) = -5 - 7t$. $f(t) = \square$ preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative of a function \( f(t) \) is \( f'(t)=\lim_{h
ightarrow0}\frac{f(t + h)-f(t)}{h} \). First, we find \( f(t + h) \) for the function \( f(t)=-5 - 7t \).

Substitute \( t+h \) into \( f(t) \): \( f(t + h)=-5-7(t + h)=-5-7t-7h \).

Step2: Compute \( f(t + h)-f(t) \)

\( f(t + h)-f(t)=(-5-7t-7h)-(-5 - 7t) \)

Simplify the expression: \( -5-7t-7h + 5 + 7t=-7h \).

Step3: Substitute into the derivative definition

Now, we substitute \( f(t + h)-f(t)=-7h \) into the limit formula:

\( f'(t)=\lim_{h
ightarrow0}\frac{-7h}{h} \)

Step4: Simplify the limit

We can cancel out the \( h \) (assuming \( h
eq0 \), which is valid since we are taking the limit as \( h
ightarrow0 \) but not equal to 0 at the moment of cancellation):

\( \lim_{h
ightarrow0}\frac{-7h}{h}=\lim_{h
ightarrow0}-7 \)

Since the function \( -7 \) is a constant, the limit as \( h
ightarrow0 \) is just \( -7 \).

Answer:

\( -7 \)