What is "A number decreased by 4 is less than 14" written as an inequality?
1 answer:
Answer:
"A number decreased by 4 is less than 14" written as an inequality is - 4 < 14
Step-by-step explanation:
First: Figure out what a number decreased by 4 is.
Well, a number decreased by 4, how would that look like? To find out, we should look for some key words in the term. So, I will bold out the keywords in the term this time:
A number decreased by 4
That word 'decreased' gives us a hint. Out of the four basic operations (addition, subtraction, multiplication and division), 'decreased' fits into the subtraction category best. So we can replace 'decreased' with a minus sign. Since we don't know the specific value of 'A number', we will assign it a variable, the variable . 4 is just 4 so we can leave that number alone. So the first term would look like this:
A number decreased by 4 = - 4
Second: Since we know what the first term is, we need to find out what the second term is and the inequality sign.
This step is quite easy. Let's tackle the inequality sign first:
A number decreased by 4 is less than 14
The only words that would describe an inequality sign are the words: 'less than'. The less than sign looks like this: <
Fourteen [just like four] is just a number so we don't need to change that
So our final product is this: - 4 < 14
Thank you for reading.
Topic: Inequalities
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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>