Respuesta :
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;
[tex]f(x) = 3x^{1/3} + 6x^{4/3}[/tex]
At the extrema points, f'(x) = 0 which gives;
[tex]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}[/tex]
(8x + 1) =x- (0/((x)^(1/0.3)) = 0
x = -1/8 = -0.125
f''(x) gives;
[tex]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}[/tex]
Substituting x = -0.125 gives f''(x) = 32 which is a minimum point
The inflection point is given as follows;
[tex]\dfrac{ \left (\dfrac{8}{3}\cdot x^2 - \dfrac{2}{3} \cdot x \right ) \times \sqrt[0.3]{x} }{x^3} = 0[/tex]
[tex]\dfrac{8}{3}\cdot x^2 - \dfrac{2}{3} \cdot x \right }{} = 0 \times \dfrac{x^3}{ \sqrt[0.3]{x}}[/tex]
[tex]\dfrac{8}{3}\cdot x - \dfrac{2}{3} \right }{} = 0[/tex]
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
