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Fantom [35]
3 years ago
14

Find the absolute maximum of m(t)=t^3-12t^2-144t and justify your answer​

Mathematics
1 answer:
ruslelena [56]3 years ago
3 0

Answer:

Step-by-step explanation:

m(t)=t³-12t²-144t

Take the derivative of t:

t'=3t²-24t-144

Set it to 0:

3t²-24t-144=0

Solve for t:

t²-8t-48=0

(t-12)(t+4)=0

t=12 or -4

On a graph, you'll find that the maximum value occurs at t=-4 ..............

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