Respuesta :
Answer: (a) Interval where f is increasing: (0.78,+∞);
Interval where f is decreasing: (0,0.78);
(b) Local minimum: (0.78, - 0.09)
(c) Inflection point: (0.56,-0.06)
Interval concave up: (0.56,+∞)
Interval concave down: (0,0.56)
Step-by-step explanation:
(a) To determine the interval where function f is increasing or decreasing, first derive the function:
f'(x) = [tex]\frac{d}{dx}[/tex][[tex]x^{4}ln(x)[/tex]]
Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),
you have:
f'(x) = [tex]4x^{3}ln(x) + x_{4}.\frac{1}{x}[/tex]
f'(x) = [tex]4x^{3}ln(x) + x^{3}[/tex]
f'(x) = [tex]x^{3}[4ln(x) + 1][/tex]
Now, find the critical points: f'(x) = 0
[tex]x^{3}[4ln(x) + 1][/tex] = 0
[tex]x^{3} = 0[/tex]
x = 0
and
[tex]4ln(x) + 1 = 0[/tex]
[tex]ln(x) = \frac{-1}{4}[/tex]
x = [tex]e^{\frac{-1}{4} }[/tex]
x = 0.78
To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:
interval x-value f'(x) result
0<x<0.78 0.5 f'(0.5) = -0.22 decreasing
x>0.78 1 f'(1) = 1 increasing
With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.
Note: As it is a natural logarithm function, there are no negative x-values.
(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:
- Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
- After 0.78, it increase (has a change of sign) and f is also defined;
Then, x=0.78 is a point of minimum and its y-value is:
f(x) = [tex]x^{4}ln(x)[/tex]
f(0.78) = [tex]0.78^{4}ln(0.78)[/tex]
f(0.78) = - 0.092
The point of minimum is (0.78, - 0.092)
(c) To determine the inflection point (IP), calculate the second derivative of the function and solve for x:
f"(x) = [tex]\frac{d^{2}}{dx^{2}}[/tex] [[tex]x^{3}[4ln(x) + 1][/tex]]
f"(x) = [tex]3x^{2}[4ln(x) + 1] + 4x^{2}[/tex]
f"(x) = [tex]x^{2}[12ln(x) + 7][/tex]
[tex]x^{2}[12ln(x) + 7][/tex] = 0
[tex]x^{2} = 0\\x = 0[/tex]
and
[tex]12ln(x) + 7 = 0\\ln(x) = \frac{-7}{12} \\x = e^{\frac{-7}{12} }\\x = 0.56[/tex]
Substituing x in the function:
f(x) = [tex]x^{4}ln(x)[/tex]
f(0.56) = [tex]0.56^{4} ln(0.56)[/tex]
f(0.56) = - 0.06
The inflection point will be: (0.56, - 0.06)
In a function, the concave is down when f"(x) < 0 and up when f"(x) > 0, adn knowing that the critical points for that derivative are 0 and 0.56:
f"(x) = [tex]x^{2}[12ln(x) + 7][/tex]
f"(0.1) = [tex]0.1^{2}[12ln(0.1)+7][/tex]
f"(0.1) = - 0.21, i.e. Concave is DOWN.
f"(0.7) = [tex]0.7^{2}[12ln(0.7)+7][/tex]
f"(0.7) = + 1.33, i.e. Concave is UP.
