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# Calculus max value of q(r)?

What is the maximum value of q(r) = r3 + 7r2 - 24r + 6 for r∈[-8, -1]?

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- Anonymous1 decade agoFavorite Answer
q(r) = r³ + 7r² - 24r + 6

q'(r) = 3r² + 14r - 24

extreme value if q'(r) = 3r² + 14r - 24 = 0

r = -6

r = 4/3

q(-6) = 186

q(-8) = 134

q(-1) = 36

conclusion, for r∈[-8, -1], q_max = 186

- 1 decade ago
Did you mean r^3 +7*r^2-24*r+6 ?

The value can reach reach maximum when differential is 0.

Differnential is 3*r^2+14*r-24.

Now 3*r^2+14*r-24=0 I hope you can solve it. then see which solution is in range r∈[-8, -1].

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