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Use the derivative to find the vertex of the parabola.
y = -x² + 4x + 2
 

Explanation 1^st Part
We need to find the derivative of                 -x² + 4x + 2
n^-1          -x²  =  -2x
n^-1        4x = 4
n^-1        2 = 0
The answer is -2x + 4
 
Explanation 2nd Part
 
We need to solve -2x + 4 = 0
First, we subtract 4 from both sides
-2x + 4 = 0 = -2x = -4
Now we divide
-4 / -2 = 2
Now we plug 2 into -x² + 4x + 2
-(2)² + 4(2) + 2 = -(4) + 8 + 2 = -4 + 8 + 2 = 4 + 2 = 6
   
For the cost and price functions​ below, find ​a) the​ number, q, of units that produces maximum​ profit;
 ​b) the​ price, p, per unit that produces maximum​ profit; and ​c) the maximum​ profit, P.
C(q) = 80 + 15q, p = 67-2q
 

 
First part
q (67 – 2q) = -2q² + 67q
Now we subtract C(q) = 80 + 15q and simplify.
-2q² + 67q – 80 – 15q = -2q² + 52q – 80
Now we find the derivative of -2q² 52q – 80
n^-1 (-2q² + 52q – 80) = -4x + 52
Now we set equal to 0 and solve
-4q + 52 = 0     add 4q to both sides  = 52 = 4q
Now we divide both sides by 4 =       q = 13
 
Second part
Nex we Plug 13 into   p = 67 – 2q
67 – 2(13) = 41
 
Third part
We subtract -2q² + 67q – 80 - 15q = -2q² + 52q – 80
Next, we plug in 13 into -2q² + 52q – 80
-2(13)² + 52(13) – 80 = 258
   
For the cost and price functions​ below, find ​the​ number, q, of units that produces maximum​ profit;
b) the​ price, p, per unit that produces maximum​ profit; and c) the maximum​ profit, P.
C(q) = 80 + 18q; p = 66 - 2q
 

 
q(66 – 2q) = -2q² + 66q
-2q² +66q – 18q – 80 =
-2q² + 48q – 80
Find derivative = -4q + 48 and solve for q
The​ number, q, of units that produces maximum profit is q = 12
 
66 – 2(12) = 42
The​ price, p, per unit that produces maximum profit is p = $42
 
-2q² + 48q – 80
-2(12)² + 48(12) – 80 =
c) The maximum profit is P = ​$208
   
For the cost and price functions​ below, find ​(​a) the​ number, q, of units that produces maximum​ profit; ​
(​b) the​ price, p, in dollars per unit that produces maximum​ profit; and (​c) the maximum​ profit, P, in dollars.

 
Explanation
q(80e^-.05) = 80q e^-.05
 
80q e^-.05q – 40 + 20q e^-.05q =    60q e^-.05q – 40
60q e^.05q – 40 - qe^.05q = 60 – 40 =        20
 
e = 2.71828
80(2.71828) ^-.05(20)
29.4304 = 29.43
60 (20) (2.71828) ^-.05(20) – 40 =           401.456 = Answer: 401.46
   
The total profit​ P(x) (in thousands of​ dollars) from a sale of x thousand units of a new product is given by
ln (-x³ + 3x² + 144x + 1) (0 ≤ x ≤ 10)
a) find the number of units that should be sold in order to maximize the total profit.
b) What is the maximum profit?
 

Find the derivative of ln (-x³ + 3x² + 144x + 1)
-3x² + 6x + 144 / -x³ + 3x² + 144x + 1 = 0
Use only the positive number 8 and multiply by one thousand = 8000
   
Suppose that the cost function for a product is given by C(x) = 0.002x³ + 7x + 6406
Find the production level​ (that is, the value of​ x) that will produce the minimum average cost per unit C(x)
 

 
The production level that produces the minimum average cost per unit is
 
Divide and simplify
.002x³ + 7x + 6406 =
.002x² + 7 + 6406 / x
Find derivative of
.002x² + 7 + 6406 / x =
.004x – 6406 / x²
Solve for x for zero (menu algebra solve)
Enter
.004x – 6406 / x² = 0,x
116.999 = 117
   
After a great deal of​ experimentation, two college senior physics majors determined that when a bottle of French champagne is shaken
several​ times, held​ upright, and​ uncorked, its cork travels according to the function​ below, where s is its height​ (in feet) above the
ground t seconds after being released.
 

 
a. Find derivative of -16t² + 30t + 4 =            -32t + 30
Solve for zero/t = 15/16 = menu number to decimal = 0.9375
Plug into the equation
-16(.9375)² + 30(.9375) + 4    = 18.0625 =     18
 
b. we need to solve the quadratic equation -16t² + 30t + 4 = 0.
 

 
a = -16², b = 30, and c = 4
 
 =      2,                     -1/8
  Click on the Degree 3 function button.  Check the Draw f’(x) box and move the​ x-slider to see f’(x) appear.
Check the Critical Points box. What do the critical points correspond​ to?
​(Use the interactive figure to find your​ answer.)
Click here to launch the interactive figure.

 
A. The critical points correspond to the zeros of f’(x)
B. The critical points correspond to the​ y-intercept.
C. The critical points correspond to where the derivative is the greatest.
   
Find the open intervals where the function is concave upward or concave downward. Find any inflection points.
 

Study the graph
   

 
Study the graph
   

Study the graph
   
Find the open intervals where the function is concave upward or concave downward. Find any inflection points.
f(x) = 3x² + 11x - 5
 

 
Find the second derivative.
1^st derivative = 6x + 11 = 2^nd derivative = 6
The second derivative is positive, so the function f is concave upward and there are no infectious points.
   
Find the open intervals where the function is concave upward or concave downward. Find any inflection points.
f(x) = -2x² – 2x + 2
 

 
Find the second derivative = -2
Since it is negative, function f is downward everywhere
https://www.emathhelp.net/calculators/calculus-1/function-calculator/
-2x^2–2x+2
enter equation then look at
Concave upward on
Concave downward on
Concave down if -∞, ∞
Then look at  Inflection Points
   
Find the open intervals where the function f(x) = is concave upward or concave downward. Find any inflection points.
 

   
Find the open intervals where the function ​f(x) = -2x³ + 12x² +169x -9
is concave upward or concave downward. Find any inflection points.
 
https://www.emathhelp.net/calculators/calculus-1/function-calculator/
enter equation

   
Find the critical point for f and then use the second derivative test to decide whether the critical point is a relative maximum or a relative minimum.
 
Find the derivative
Then solve for zero and x
Then plug the value into the original equation
Value of x and the answer are the critical points
Then follow the second derivative test.
The second derivative is  -2
 
The second derivative test states the following.

 
 
Find any critical numbers for the function f(x) and then use the second derivative test to decide whether the critical
numbers lead to relative maxima or relative minima. If the second derivative test gives no​ information, use
the first derivative test instead.
 
https://www.emathhelp.net/calculators/calculus-1/function-calculator/
find derivative first
then enter the equation and go to
x-intercepts
use the bottom
maximum at smallest, minimum at largest

 
https://www.emathhelp.net/calculators/calculus-1/function-calculator/
 
Find the point of diminishing returns (x,y) for the function​ R(x), where​ R(x) represents revenue​ (in thousands of​ dollars) and x represents the amount spent on advertising​ (in thousands of​ dollars).
 
Find second derivative then
go to menu algebra solve for zero and x.
then plug the answer into the equation
 
For the function f(x) = 3x³ -8x² + 13x + 4. find f”(x) then find f”(0) and f”(3)
 
f”(x) means derivative.
Find the second derivative and plug in the values (0) and (3)
 
 

For the function f(x) = x² / 2 + x find f”(x) then find f”(0) and f”(3)
 
Find the second derivative
Then solve the quadratic equation.
Plug both solutions in and use the largest of the two.
Then add on to the year

Homework: Section 5.4 Graph the​ function, considering the​ domain, critical​ points, symmetry, relative​ extrema, regions where the function is increasing or​ decreasing,
inflection​ points, regions where the function is concave upward or concave​ downward, intercepts where​ possible, and asymptotes where applicable.
 
First find first and second derivatives.
Then go to
https://www.emathhelp.net/calculators/calculus-1/function-calculator/
enter the function and use the local minima and maxima.
Local Minima
(−4,−844)
 
Local Maxima
(3,528)
 

 
Local Minima
(−5,−289)
Local Maxima
(2,54)
 
Graph the​ function, considering the​ domain, critical​ points, symmetry, regions where the function is increasing or​ decreasing, inflection​ points,
regions where the function is concave upward or concave​ downward, intercepts where​ possible, and asymptotes where applicable.
 
https://www.emathhelp.net/calculators/calculus-1/function-calculator/
enter the function and use the local minima and maxima.
 
Local Minima
(−4,−636)
Local Maxima
(3,393)

Graph the​ function, considering the​ domain, critical​ points, symmetry, regions where the function is increasing or​ decreasing, inflection​ points,
regions where the function is concave upward or concave​ downward, intercepts where​ possible, and asymptotes where applicable.
 
 
https://www.emathhelp.net/calculators/calculus-1/function-calculator/
enter the function and use the local minima and maxima.
 
Local Minima
(0,−2812)
Local Maxima
No maxima.
 
 

 
Local Minima
(0,−2587)
Local Maxima
No maxima.
 

 
Graph the​ function, considering the​ domain, critical​ points, symmetry, regions where the function is increasing or​ decreasing, inflection​ points,
regions where the function is concave upward or concave​ downward, intercepts where​ possible, and asymptotes where applicable.
 
 
https://www.emathhelp.net/calculators/calculus-1/function-calculator/
enter the function and use the local minima and maxima.
 
Local Minima
(0,−2812)
Local Maxima
No maxima.
 

   
 
Graph the​ function, considering the​ domain, critical​ points, symmetry, relative​ extrema, regions where the function is increasing or​ decreasing,
inflection​ points, regions where the function is concave upward or concave​ downward, intercepts where​ possible, and asymptotes where applicable.
 
Find the first and second derivatives with the ti-nspire.
Then go to
https://www.emathhelp.net/calculators/calculus-1/function-calculator/
enter the function and go to the bottom.
Graph
For graph, see graphing calculator.
 
 
 
 
The graph of a​ company's profit​ P(t) in​ dollars, at month t is shown. Complete parts a through e below.


Study the graph
  The graph of a​ company's profit​ P(t) in​ dollars, at month t is shown.
Complete parts a and b below.
 


 
Study the graph
   
The graph of a company’s profit P(t) in dollars, at month t is shown
Complete parts a through e below
 


Study the graph
   
The graph of a​ company's profit​ P(t) in​ dollars, at month t is shown.
Complete parts a and b below.
 

Study the graph
   
The graph of a​ company's profit​ P(t) in​ dollars, at month t is shown.
Complete parts a and b below.
 

Study the graph
   
The graph of a​ company's profit​ P(t) in​ dollars, at month t is shown.
Complete parts a and b below.
 

Study the graph
   
The daily​ demand, q, for cupcakes is a function of the price​ p, where
q = f(p) = 150 – 39p
 

a. plug in the value of f(p) = 1
f(p) = 150 – 39(1) = 111
 
b. we need to find the derivative and then plug in the value of f’(1)        *f’ = derivative
f’ = 150 – 39p = -39(1) = 39
   
The daily​ demand, q, for cupcakes is a function of the price​ p, where
q = f(p) = 110 – 28p
 

 
a. plug in the value of f(p) = 2
f(p) = 110 – 28(2) = 54
 
b. we need to find the derivative and then plug in the value of f’(2)        *f’ = derivative
f’ = 110 – 28p = -28(2) = 28

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