The population of fish in a lake could be modeled by the function f(t)= 40t/(t^2+1), where t is given in days. The function that actually models the fish population is g(t)=45t/(t^2+8t+7). Determine where g(t)>f(t).

2. Originally Posted by BuffaloSoulja
The population of fish in a lake could be modeled by the function f(t)= 40t/(t^2+1), where t is given in days. The function that actually models the fish population is g(t)=45t/(t^2+8t+7). Determine where g(t)>f(t).
you need to solve $\displaystyle \frac {45t}{t^2 + 8t + 7} > \frac {40t}{t^2 + 1}$

Bring everything to the left side, combine, simplify and set both the numerator and denominator equal to zero. Solve for the t-values. Plot those on a number line, and test the intervals to see which ones you need.

can you finish?

3. Originally Posted by BuffaloSoulja
The population of fish in a lake could be modeled by the function f(t)= 40t/(t^2+1), where t is given in days. The function that actually models the fish population is g(t)=45t/(t^2+8t+7). Determine where g(t)>f(t).
$\displaystyle g(t)>f(t)$

$\displaystyle \frac{45t}{t^2+8t+7} > \frac{40t}{t^2+1}$

$\displaystyle 45t(t^2+1) > 40t(t^2+8t+7)$

$\displaystyle 45t^3+45t > 40t^3+320t^2+280t$

$\displaystyle 5t^3-320t^2-235t > 0$

Can you solve it from here?

4. Originally Posted by pickslides
$\displaystyle g(t)>f(t)$

$\displaystyle \frac{45t}{t^2+8t+7} > \frac{40t}{t^2+1}$

$\displaystyle 45t(t^2+1) > 40t(t^2+8t+7)$

$\displaystyle 45t^3+45t > 40t^3+320t^2+280t$

$\displaystyle 5t^3-320t^2-235t > 0$

Can you solve it from here?

Alright. I got to $\displaystyle 5t(t^2-64t-47)/(t+1)(t+7)(t^2+1) >0$

I can just cancel out the denominator? Would I not lose solutions?

5. Originally Posted by BuffaloSoulja
Alright. I got to $\displaystyle 5t(t^2-64t-47)/(t+1)(t+7)(t^2+1) >0$

I can just cancel out the denominator? Would I not lose solutions?
I get

$\displaystyle 5t(t^2-64t-47) >0$

not sure how you found those terms in the denominator.

6. Originally Posted by pickslides
I get

$\displaystyle 5t(t^2-64t-47) >0$

not sure how you found those terms in the denominator.

$\displaystyle g(t)>f(t)$
$\displaystyle 45t/(t^2+8t+7)>40t/(t^2+1)$
$\displaystyle 45t/(t+1)(t+7)-40t/(t^2+1)>0$
Find LCD by multiplying 1
$\displaystyle 45t/(t+1)(t+7) * (t^2+1)/(t^2+1)-40t/(t^2+1) * (t+7)(t+1)/(t+7)(t+1) > 0$
Simplifies to
$\displaystyle (5t^3-320t^2-235t)/(t+1)(t+7)(t^2+1)$
$\displaystyle 5t(t^2-64t-47)/(t+1)(t+7)(t^2+1) >0$

This is what i did btw