anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\frac{\frac{1}{t}-1}{t-1}=-15\] solve for \(t\)
anonymous
  • anonymous
Average rate of change: \[ \overline{f'(t)} = \frac{f(t_2)-f(t_1)}{t_2-t_1} \]
anonymous
  • anonymous
what @wio sain in your case \(t_1=1\) and so \(f(t_1)=f(1)=1\) also \(t_2=t\)

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anonymous
  • anonymous
*said
anonymous
  • anonymous
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
  • anonymous
Multiply both sides by \(t\).
anonymous
  • anonymous
what do you mean by multiply both sides by t?
anonymous
  • anonymous
\[ 1 = -\frac 1 5t^2+\frac 65 t \]
anonymous
  • anonymous
\[ 5 = -t^2+6t \]
anonymous
  • anonymous
This is just a quadratic equation.
anonymous
  • anonymous
Makes sense thank you!

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