Question

Determine the intervals on which the following function is concave up or concave down. Identify any inflection points. g (t )equals3 t Superscript 5 Baseline minus 10 t Superscript 4 Baseline minus 30 t cubed plus 110

Answer 1

Answer:

The inflection points are t=0, t=3 and t=-1

For (-∞,-1): concave down

For (-1,0): concave up

For (0,3): concave down

For (3,∞): concave up

Step-by-step explanation:

The function is: $g(t)=3t^5-10t^4-30t^3+110$

First, we need to calculated the second derivate $g''(t)$ as:

$g'(t)=15t^{4}-40t^{3}-90t^{2} \\g''(t)=60t^{3}-120t^{2}-180t$

Now, the inflection points are the points in which the second derivate is equal to zero, so, the inflection points are calculated as:

$g''(t)=60t^{3}-120t^{2}-180t=0\\t(60t^{2}-120t-180)=0\\t=0\\ or\\60t^{2}-120t-180=0\\t^{2}-2t-3=0\\(t-3)(t+1)=0\\t=3\\or\\t=-1$

Then, the inflection points are t=0, t=3 and t=-1, so we have the following intervals: (-∞,-1) (-1,0) (0,3) and (3,∞)

Thus, if the second derivate is positive in the interval, the function is concave up and if the second derivate is negative in the interval, the function is concave down. So, for every interval we have:

For (-∞,-1)

$g''(-2)=60(-2)^{3}-120(-2)^{2}-180(-2)=-600$

For (-1,0)

$g''(-0.5)=60(-0.5)^{3}-120(-0.5)^{2}-180(-0.5)=52.5>0$

For (0,3)

$g''(1)=60(1)^{3}-120(1)^{2}-180(1)=-240$

For (3,∞)

$g''(4)=60(4)^{3}-120(4)^{2}-180(4)=1200>0$

It means that the function is:

For (-∞,-1): concave down

For (-1,0): concave up

For (0,3): concave down

For (3,∞): concave up