How do you find the instantaneous rate of change of g(t)=3t2+6 at t=4?

Question

tigging9k0 · Accepted Answer

Explanation:Compute the first derivative:g&#039;(t)=6tEvaluate it at t=4g&#039;(4)=24

Alice Harmon · Accepted Answer

If you have learned the power rule, constant multiple rule and derivative of a constant, you can quickly find the derivative of g.

g^{'} (t) = 3 (2 x^{1}) + 0 = 6 t

.
To find the instantaneous rate of change at a particular value of t, evaluate the derivative at that value of t.
At t=4 the instantaneous rate of change is g'(4)=6(4)=24.
If you are using a definition then it depends on the particular definition you are using.
There are several ways to express the definition.
One way of expressing it is to give:
The rate of change of g with respect to t at t=4 is

lim_{t \to 4} \frac{g (t) - g (4)}{t - 4}

Another is
The rate of change of f with respect to t at t=4 is

lim_{h \to 0} \frac{g (t + h) - g (t)}{h}

After we find this, we evaluate at t=4.
Here is the work for the first definition above.

lim_{t \to 4} \frac{g (t) - g (4)}{t - 4} = lim_{t \to 4} \frac{[3 t^{2} + 6] - [3 {(4)}^{2} + 6]}{t - 4}

(Observe that is we substitute t=4, we get the indeterminate form

\frac{0}{0}

.)

= lim_{t \to 4} \frac{3 t^{2} + 6 - 48 - 6}{t - 4}

(Still

\frac{0}{0}

)

= lim_{t \to 4} \frac{3 t^{2} - 48}{t - 4}

= lim_{t \to 4} \frac{3 (t^{2} - 16)}{t - 4}

= lim_{t \to 4} (3 (t + 4) (t - 4)) \frac{1}{t - 4}

= lim_{t \to 4} 3 (t + 4)

=3(4+4)=24

How do you find the instantaneous rate of change of

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