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zmey [24]
3 years ago
15

H(x)=1/8x^3-x^2 Over which interval does h have a positive average rate of change?

Mathematics
1 answer:
lara31 [8.8K]3 years ago
6 0

Answer:

(-\infty,0)\cup(\frac{16}{3},\infty)\\That \ is x\frac{16}{3}

Step-by-step explanation:

h(x)=\frac{1}{8}x^{3} -x^{2} \\\\Differentiate \ h(x) \ with \  respect \ to \ x\\h'(x)=\frac{3}{8}x^{2} -2x

For positive rate of change h'(x)>0

\frac{3}{8} x^{2} -2x>0\\x(\frac{3}{8}x-2)>0\\\\ When \ x0 \ (multiplication \ of \ two \ positives \ is \ positive)\\\\h'(x)>0 \ when \ x\frac{16}{3} \\\\

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