Let f(t)=(1/t). Find a value of t such that the average rate of change of f(t) from 1 to t equals -15 .

With the t do times the t and (1/t) so it's like (1/t^2)? Confused.

Question

Let f(t)=(1/t). Find a value of t such that the average rate of change of f(t) from 1 to t equals -15 . 

With the t do times the t and (1/t) so it's like (1/t^2)? Confused.

anonymous · Answer

$$\frac{\frac{1}{t}-1}{t-1}=-15$$ solve for $t$

anonymous · Answer

Average rate of change: $$
\overline{f'(t)} = \frac{f(t_2)-f(t_1)}{t_2-t_1}
$$

anonymous · Answer

what @wio  sain
in your case $t_1=1$ and so $f(t_1)=f(1)=1$ 
also $t_2=t$

anonymous · Answer

*said

anonymous · Answer

oh okay! my apologies but -15 is actually -1/5. I typed it in wrong. So I am now stuck at this part (1/t)=(-1/5)t+(6/5) . I know you bring over everything to one side and factor but how do factor the (1/t)? Or I am I completely way off?

anonymous · Answer

Multiply both sides by $t$.

anonymous · Answer

what do you mean by multiply both sides by t?

anonymous · Answer

$$
1 = -\frac 1 5t^2+\frac 65 t
$$

anonymous · Answer

$$
5 = -t^2+6t
$$

anonymous · Answer

This is just a quadratic equation.

anonymous · Answer

Makes sense thank you!