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.
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
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