Calculus 1 problem - derivatives
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Calculus 1 problem - derivatives

[From: ] [author: ] [Date: 11-11-15] [Hit: ]
1547,Therefore,......
f(x) = (x^3−4x)^1/3

Indicate the transition points (local extrema and points of inflection)

I know I need to find the critical points by setting the first derivative equal to 0 then plugging those into the original function to find the local extrema. The trouble I'm really having is taking the second derivative and setting that equal to 0 to solve that for x to find the points of inflection.

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f(x) = (x³−4x)¹ʹ³

f '(x) = ⅓(x³−4x)⁻²ʹ³(3x²-4) = 0 only for 3x²-4=0   ⇒   x = ±2/√3  ≈  ±1.1547
But, f is undefined at x = 2/√3. So, only one critical point at x = -2/√3     ← ANSWER   ➊

Now, get the 2nd derivative.
f '(x) = ⅓(x³−4x)⁻²ʹ³(3x²-4)            ← f ' = ⅓ uv
f ''(x) = ⅓ [(x³−4x)⁻²ʹ³•(6x-0) + (3x²-4)•(-⅔(x³−4x)⁻⁵ʹ³)(3x²-4)
f ''(x) = ⅓ [6x(x³−4x)⁻²ʹ³ - ⅔(3x²-4)(x³−4x)⁻⁵ʹ³(3x²-4)]         ← now, factor out (x³−4x)⁻²ʹ³
f ''(x) = ⅓ (x³−4x)⁻²ʹ³[6x - ⅔(3x²-4)²(x³−4x)⁻¹ ]                  ← not pretty, but it will suffice

           ⅓ (x³−4x)⁻²ʹ³[6x - ⅔(3x²-4)²(x³−4x)⁻¹ ] = 0
                    Possible solutions are for only
                      [6x - ⅔(3x²-4)²(x³−4x)⁻¹ ] = 0
                              - ⅔(3x²-4)²(x³−4x)⁻¹ = -6x
                                   (3x²-4)²(x³−4x)⁻¹ = 9x
                                                (3x²-4)² = 9x(x³−4x)
                                         9x⁴-24x²+16 = 9x⁴-36x²                                     ANSWER
                                                         x² = -⁴∕₃        ← no solution   ⇒   no inflection points


Now, from ➊, all that is left to determine is whether there is a local max or min
at x = -2/√3 ... i.e. is f" negative or positive?


Doing the algebra to simplify the 2nd derivative gets us:
             -(24x² + 32)            ← notice that the numerator is negative for all x
f ''(x) = ——————
             9∛[(x³-4x)⁵]            ← the sign of the denominator is determined by x³-4x.
                                                       For x = -2/√3  ≈  -1.1547, (x³-4x) is positive
                                                       so that the denominator of f '' is positive

Therefore, f ''( -2/√3) = neg/pos = neg ⇒ There is a local max at  x =  -2/√3 ←ANSWER
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