Question

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 3x(1/3) + 6x(4/3). You must justify your answer using an analysis of f ′(x) and f ′′(x).

Answer 1

We have an extrema (local minimum) at x = -0.125

An inflection point at x = 0.25

Explanation:

The given function is given as follows;

`f(x) = 3x^(1/3) + 6x^(4/3)`

At the extrema points, f'(x) = 0 which gives;

`0 = \frac{\mathrm{d} \left (3x^(1/3) + 6x^(4/3) \right )}{\mathrm{d} x} = \frac{(8 \cdot x+1) * \sqrt[0.3]{x} }{x}`

(8x + 1) =x- (0/((x)^(1/0.3)) = 0

x = -1/8 = -0.125

f''(x) gives;

`f''(x) = \frac{\mathrm{d} \left (\frac{(8 \cdot x+1) * \sqrt[0.3]{x} }{x} \right )}{\mathrm{d} x} = \frac{ \left ((8)/(3)\cdot x^2 - (2)/(3) \cdot x \right ) * \sqrt[0.3]{x} }{x^3}`

Substituting x = -0.125 gives f''(x) = 32 which is a minimum point

The inflection point is given as follows;

`\frac{ \left ((8)/(3)\cdot x^2 - (2)/(3) \cdot x \right ) * \sqrt[0.3]{x} }{x^3} = 0`

`(8)/(3)\cdot x^2 - (2)/(3) \cdot x \right }{} = 0 * \frac{x^3}{ \sqrt[0.3]{x}}`

`(8)/(3)\cdot x - (2)/(3) \right }{} = 0`

x = 2/3×3/8 = 1/4 = 0.25

We check the value of f''(x) at x = 0.24 and 0.26 to determine if x = 0.25 is an inflection point as follows;

At x = 0.24, f''(x) = -0.288

At x = 0.26, f''(x) = 0.252

0.25 is an inflection point