Please help!!!!! In a hurry
2 answers:
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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, ∞).
Answer:
restaurant should charge $(6-0.25) = $5.75per sandwich to maximize daily revenue.
the revenue is $1983.75
Step-by-step explanation:
to calculate current revenue= $6 x 330 = $1980
suppose x as the number of times the price to be dropped by $0.25
then find new price.. i.e
new price= $(6-0.25x)
and, new sell=330 +15x sandwiches
therefore, the new revenue would be= (6-0.25x)(330 +15x)
in order to maximize the current revenue, simplify the above equation and make it complete square using x
(6-0.25x)(330 +15x)
=1980-82.5x +90x -3.75
=1980 + 7.5x -3.75
=1980-3.75 (-2x+) ----> taking out common
now, to make a complete square lets add and subtract 1 inside the parentheses
=1980-3.75(-1+1-2x+)
=1980 +3.75 -3.75( -2x +1)
=1983.75 -3.75 ---->(1)
as is positive always, minimize the other term in order to maximize the total revenue.
so the minimum possible value of = 0
therefore, x=1
putting x in eq(1) the revenure becomes,
$(1983.75-0)=> $1983.75
therefore, restaurant should charge $(6-0.25) = $5.75per sandwich to maximize daily revenue.
the revenue is $1983.75