Write an algebraic expression that models the situation.
2 answers:
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A coordinate plane with the x-axis labeled Blueberries (in pints) and the y-axis labeled Cost (in dollars) shows a line going through (0, 0) and (5, 17).
<h3>What is directly proportional?</h3>It is defined as the relationship between two quantities as one quantity increases the other quantity also increases and vice versa.
It is given that there is a proportional relationship between cost and pints of blueberries as,
Blueberries (in pints) 2 5
Cost (in dollars) 6.80 17
17/6.80 = 5/2
As we draw the graph and analyze each option one by one we get the graph shown by the statement C following the given proportional relationship.
Thus, a coordinate plane with the x-axis labeled Blueberries (in pints) and the y-axis labeled Cost (in dollars) shows a line going through (0, 0) and (5, 17).
Learn more about the directly proportional here:
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Answer:
(a) Decreasing on (0, 1) and increasing on (1, ∞)
(b) Local minimum at (1, 0)
(c) No inflection point; concave up on (0, ∞)
Step-by-step explanation:
ƒ(x) = x² - x – lnx
(a) Intervals in which ƒ(x) is increasing and decreasing.
Step 1. Find the zeros of the first derivative of the function
ƒ'(x) = 2x – 1 - 1/x = 0
2x² - x -1 = 0
( x - 1) (2x + 1) = 0
x = 1 or x = -½
We reject the negative root, because the argument of lnx cannot be negative.
There is one zero at (1, 0). This is your critical point.
Step 2. Apply the first derivative test.
Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.
(1) x = ½
ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1
ƒ'(x) < 0 so the function is decreasing on (0, 1).
(2) x = 2
ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½ = ⁵/₂
ƒ'(x) > 0 so the function is increasing on (1, ∞).
(b) Local extremum
ƒ(x) is decreasing when x < 1 and increasing when x >1.
Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.
(c) Inflection point
(1) Set the second derivative equal to zero
ƒ''(x) = 2 + 2/x² = 0
x² + 2 = 0
x² = -2
There is no inflection point.
(2). Concavity
Apply the second derivative test on either side of the extremum.
The function is concave up on (0, ∞).
