I need help solving this problem, If f(x)=x^3+ax^2+bx+c Solve for a and b if there is a local max at x=1 and point of inflection x=3.?
I don’t know how to solve for the constants when the POI is known and local max is known.
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answers:
TomV say: Solve for a and b if there is a local max at x=1 and point of inflection x=3

Local maximum at x = 1 requires that f'(1) = 0 and f"(1) < 0
Point of inflection at x = 3 requires that f"(3+ε) = 0 for ε = 0 and f"(3+ε)•f"(3-ε) < 0 for |ε| << 1

f(x)=x^3+ax^2+bx+c
f'(x) = 3x² + 2ax + b
f"(x) = 6x + 2a

local max at x=1
f'(1) = 0 = 3 + 2a + b → b = -(2a+3)
f"(1) = 6 + 2a < 0 → a < -3

point of inflection x=3
f"(3) = 0 = 18 + 2a → a = -9 : meets requirement that a < -3

b = -(-18+3) = 15

f(x) = x³ - 9x² + 15 + c
f'(x) = 3x² - 18x + 15
f"(x) = 6x - 18

f"(3+ε) = 6(3+ε) - 18 = 6ε
f"(3+ε) > 0 if ε > 0
f"(3+ε) < 0 if ε < 0
Since f"(x) changes sign on [3-ε, 3+ε] for |ε| << 1, x = 3 is a point of inflection.

Ans:
a = -9
b = 15
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rotchm say: The idea is the following:

You are told that f ' (1) = 0 and f ''(3) = 0.
So, just find these derivatives.
This generates a linear system of two unknowns a & b. Just solve for a & b as you saw in highschool.
Done!

Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them.
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david say: f(x)=x^3+ax^2+bx+c
f' = 3x^2 + 2ax + b = 0 at x = 1
3 + 2a + b = 0
f'' = 6x + 2a= 0 at x = 3
a = -9
from f' ... b = 15
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Anonymous say: Ok
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