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Maximum and minimum values

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Given: y = f(x) = 3x4 + 4x3

A. All critical points
B. Max - Min Values
C. Inflection points
D. Where is f(x) concave up
E. Where is f(x) concave down
F. X and Y intercepts
G. Where f(x) is increasing
H. Where f(x) is decreasing
I. Sketch the curve label

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Solution Summary

This shows how to find critical points and concavity.

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f(x) = 3x^4 + 4x^3
=> f'(x) = 12x^3 + 12x^2 = 12x^2(x+1)
=> f''(x) = 36x^2 + 24x
=> f'''(x) = 72x + 24
=> f''''(x) = 72
For critical points:
f'(x) = 0 => 12x^2(x+1) = 0
=> x = 0, -1 --Answer

=>f''(0) = 0 : Neither max nor min
=> f'''(0) =
f''(-1) = 36 - 24 = 12: minima

f(infinity) = infinity
f(-infinity) ...

Solution provided by:
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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