How Do You Find A and B for an Inflection Point in a Function?

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The discussion focuses on determining the constants A and B in the function y = Ax^(1/9) + Bx^(-1/9) to achieve an inflection point at (1, 9). The key equations derived include A + B = 9 and -8A + 10B = 0. The incorrect calculations led to A = -5 and B = 14, highlighting algebraic errors in sign management. Correcting these mistakes is essential for accurate results in calculus problems involving inflection points.

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Homework Statement


Determine A and B so that y = Ax^(1/9) + Bx(-1/9) has an inflection point as (1/9)

Homework Equations



A + B = 9

The Attempt at a Solution



F' = Ax^(-8/9)/9 - Bx^(-10/9)/9
F'' = -8Ax^(---)/81 + 10Bx^(---)/81

-8A + 10B = 0
A + B = 9
-10(9-A) - 8A = 0
-90 - 18A = 0
A = -90/18
B = 14

But, this is not correct.. I don't understand what I did wrong?
 
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Chas3down said:

Homework Statement


Determine A and B so that y = Ax^(1/9) + Bx(-1/9) has an inflection point as (1/9)

Homework Equations



A + B = 9

The Attempt at a Solution



F' = Ax^(-8/9)/9 - Bx^(-10/9)/9
F'' = -8Ax^(---)/81 + 10Bx^(---)/81

-8A + 10B = 0
A + B = 9
-10(9-A) - 8A = 0
-90 - 18A = 0
A = -90/18
B = 14

But, this is not correct.. I don't understand what I did wrong?
I'm assuming you meant "at the point (1,9)" not "as (1/9)".

Check your algebra when solving for A and B. You made a few sign mistakes.
 

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