Practice working with reflections.
2 answers:
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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) = [
]
Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),
you have:
f'(x) =
f'(x) =
f'(x) =
Now, find the critical points: f'(x) = 0
= 0
x = 0
and
x =
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) =
f(0.78) =
f(0.78) = - 0.092
The point of <u>minimum</u> 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) = [
]
f"(x) =
f"(x) =
= 0
and
Substituing x in the function:
f(x) =
f(0.56) =
f(0.56) = - 0.06
The <u>inflection point</u> 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) =
f"(0.1) =
f"(0.1) = - 0.21, i.e. <u>Concave</u> is <u>DOWN.</u>
f"(0.7) =
f"(0.7) = + 1.33, i.e. <u>Concave</u> is <u>UP.</u>