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

Answer:

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

An inflection point at x = 0.25

Step-by-step 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;

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

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

x = -1/8 = -0.125

f''(x)  gives;

$f''(x) = \dfrac{\mathrm{d} \left (\dfrac{(8 \cdot x+1) \times \sqrt[0.3]{x} }{x} \right )}{\mathrm{d} x} = \dfrac{ \left (\dfrac{8}{3}\cdot x^2 - \dfrac{2}{3} \cdot x \right ) \times \sqrt[0.3]{x} }{x^3}$

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

The inflection point is given as follows;

$\dfrac{ \left (\dfrac{8}{3}\cdot x^2 - \dfrac{2}{3} \cdot x \right ) \times \sqrt[0.3]{x} }{x^3} = 0$

$\dfrac{8}{3}\cdot x^2 - \dfrac{2}{3} \cdot x \right }{} = 0 \times \dfrac{x^3}{ \sqrt[0.3]{x}}$

$\dfrac{8}{3}\cdot x - \dfrac{2}{3} \right }{} = 0$

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