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Find all the critical numbers of the function.
f(x) = x³ - x² – 15x + 2
 

 
Use the first numbers in (x,f(x)).
To find the critical numbers of a function the long way we must find the derivative,
set the derivative to equal zero (0), and solve for x.
 
First find the derivative of:
f(x) = x³ - x² – 15x + 2
3(x^{3 – 1}) = 2x²
2(- x^{2 - 1}) = -x
1(-15x1^-1) = 15
2 ( ^{- 1}) = 0
Derivative is: 2x² – x – 15
Then set the derivative to equal zero (0).
2x² – x – 15 = 0
Solve quadratic equation
http://coursesavior.com/free/quadratic-equation-solver/
3, -2.5 (or -5/2)
 
This calculator will find the critical points. Enter in precisely as shown and scroll down to
Critical Points
https://www.emathhelp.net/calculators/calculus-1/function-calculator
 

   
Find all the critical numbers of the function.
f(x) =
 

To find the critical numbers of a function the long way we must find the derivative,
set the derivative to equal zero (0), and solve for x.
 
The derivative = 0, therefore there are no Critical Points.
Let’s simplify the equation to see for ourselves!
f(x) =
 
To simplify we would do so by:
x – (x) = 0
-4 – (7) = -11
The derivative of -11 = 0 therefore there are no critical points!
 
This calculator will find the critical points. Enter in precisely as shown and scroll down to
Critical Points
https://www.emathhelp.net/calculators/calculus-1/function-calculator

Scroll Down to Critical Points
No critical points.
 
https://www.derivative-calculator.net/
 

  Find the open​ interval(s) where the function is changing as requested.
    y = x⁴ – 18x² + 81
 

The long way.
To find the Intervals of Increase of this function the long way we must find the derivative,
set the derivative as so, and solve for x.
Let’s find the derivative of x⁴ – 18x² + 81.
n^{n –1}                 4(x^{4 – 1}) = 4x³                    2(-18^{2 – 1}) = -36x              81(0) = 0
Derivative =                            4x³ – 36x
Then we set up the equation. Since we are looking for the intervals of increase,
we must use the greater than sign (>) and set at zero (0)
4x³ – 36x > 0   factored = 4x (x² – 9) = 0
Then we must factor some more and set each set to equal zero (0)
4x(x² – 3²) = 0              4x(x + 3)(x – 3) = 0
So, we have 4x(x + 3)(x – 3) = 0
What we can do next is set each to equal zero.
4x = 0              x + 3 = 0           x – 3 = 0
4x = 0 =            0
x + 3 = 0 =        -3
x – 3 = 0 =        3
the answer = (-3,0) U (3,∞)
 
https://www.emathhelp.net/calculators/calculus-1/function-calculator
Intervals of Increase

 
Scroll Down to:
Find the location and value of all relative extrema for the function.
 

 
 
Explanation!
A function has relative minimum at x = a, if f'(a) = 0 and f" (a) > 0 (concave upward)
and Relative maximum at x = b, if f'(b) = 0 and f" (b) <0 (concave downward).
Find the location and value of all relative extrema for the function.


Explanation!
A function has relative minimum at x = a, if f'(a) = 0 and f" (a) > 0 (concave upward)
and Relative maximum at x = b, if f'(b) = 0 and f" (b) <0 (concave downward).
The annual revenue and cost functions for a manufacturer of grandfather clocks are approximately
R(x) = 500x - .03x² and C(x) = 160x + 100,000 where x denotes the number of clocks made.
What is the maximum annual​ profit?
 

 
This is a little tricky, but we solve it like so…
Values:
R(x) = 500x - .03x² and C(x) = 160x + 100,000
 
Gather values:                        500x - .03x² – 160x – 100,000.
Simplify:                                 500x - .03x² – 160x – 100,000 = 340x - .03x² – 100,000
Find derivative                       340x - .03x² – 100,000 = 340 - .06x (*We need to plug in final value into this derivative)
Solve   for x                             340 - .06x = 5666.67

plug in to equation    340(5666.67) - .03(5666.67)² – 100000 = 863,333
   
Find the number of​ units, x, that produces the maximum profit​ P,
if C(x) = 40 + 52x and p = 84 – 2x.
 

 
First set up the equation
x (84 – 2x) – 40 – 52x
Simplify          -2x²+ (84x – 52x) – 40            simplify          -2x² + 32x – 40
Find the derivative of -2x² + 32x – 40.
2(2x^{2 – 1}) =        -4x
1(32x^{1 – 1}) =      32
-40(0) =           0
Derivative = -4x + 32
Now set the equation to 0 and solve for x.
-4x + 32 = 0
Add 4x to both sides.             32 = 4x
Divide both sides by 4.           x = 8
  Find the largest open interval where the function is concave upward.
            f(x) = x² + 2x + 1

To solve this problem, we need to find the Domain, however there is no Domain,
So, we use Interval Notation: (−∞, ∞).
Use the functions calculator.
 
https://www.emathhelp.net/calculators/calculus-1/function-calculator
 

 
Scroll down to
 
 
The reason there is no Domain, and determined Interval Notation is because
there is no Real Number to make the expression undefined.
 
Decide if the given value of x is a critical number for​ f, and if​ so, decide whether the point is a relative​ minimum,
relative​ maximum, or neither.
 
f(x) = x⁵; x = 0
 

To solve this problem, we can use the first or second derivatives or not.
 We can set both equations to zero, solve for x, and the critical number will be 0.
However, there is no extreme point because there are no ordered pairs.
 
x⁵ = 0
1^ST Derivative: 5x⁴ = 0             =          0
2^ND Derivative:  20x³ = 0         =          0


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.
f(x) = -6x³ – 36x² + 216x - 11

 
 
Explanation

It takes a little practice, but you can look at all the other graphs, and see the above is the correct graph.
To solve this problem, we need to find the Local Minima
This calculator will find and compare those to the graphs.
https://www.derivative-calculator.net/
In this case, Local Minima are (-6, -1307)
You can find the Local Minima below. Enter the equation as shown below. Solve this problem the long way.
Find the first derivative, set at zero, and solve for x.
I won’t put all the steps to finding the derivatives because you should know how to do that by now.
Or use this calculator.
https://www.derivative-calculator.net/
 
Find the derivative of:           -6x³ – 36x² + 216x – 11.
            First Derivative           -18x² – 72x + 216
Set to equal 0 and solve for x.     -18x² – 72x + 216 = 0
We need to factor -18x² – 72x + 216             This is from Algebra 1, so I will not show the steps.
-18x² – 72x + 216 factored out =       -18(x – 2) (x + 6)
Next, we do what we did above
-18 = 0 =          0                     
x – 2 = 0 =        2                     
x + 6 = 0 =        -6
The first local minima we need would be (-6)
To find the second local maxima we need to find the “y value” when x = -6
This is true because the value of the second derivative is positive.
All this means is we plug in the value -6 to the ENTIRE equation,  which means f(x).
*Just a note, you would not need to go any further, because none of the other graphs
represented a “local maximum” of “-6”, however, we will solve it anyway.
 
f (x) = -6x³ – 36x² + 216x – 11
f (-6) = -6(-6)³ – 36(-6)² + 216(-6) – 11
f (−6) = 1296 – 1296 – 1296 – 11
f (−6) = −1307 (We do not need to divide by -6 in this case, because we do not need f to stand alone)
Local Maxima = (-6, -1307)
Now we can compare the graphs and get the answer right!
 
https://www.emathhelp.net/calculators/calculus-1/critical-points-extrema-calculator/
 
 

   
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.
f(x) = x⁴ – 54x² + 245
 

 
Explanation: Graph with points

To solve this problem, we need to find the Local Minima and the Local Maxima
This calculator will find and compare those to the graphs.
 (-5.196, -484) (0, 245) (5.196,484)
 
You can find the Local Minima below. Enter the equation as shown below.
https://www.emathhelp.net/calculators/calculus-1/critical-points-extrema-calculator/


The long way
To find the local maxima, we need to find the derivative of the given equation. f(x) = x⁴ – 54x² + 245.
Again, after you start calculating derivatives, you will be able to do them quite easily.
 
I can do this by eye.
x⁴ = 4x³            - 54x² = -108x             245 = 0         the derivative = 4x³ – 108x
Next step is:
Solve for x: 4x³ – 108x = 0
Answer: x = 0 or x = 5.196152422706632 or x = −5.196152422706632
We can now see that one of the points to plot is zero (0). Two of the graphs in question
Have points plotted equal at 0. We are already given the y-value of 245 however, one of the graphs
has values of 0, 245 and 0, -245, so we will find the y-value when x = 0.
Plug in the x value of 0 into
f(x) = x⁴ – 54x² + 245               f (0) = (0)⁴ – 54(0)² + 245 =     245
The Local Maxima is (0, 245) and only one of the graphs have those values plotted,
so, we do not need to go any further and do not need to find the local minima.
However, if we wanted to all we need to do is plug in one of the values we have already
found which is (5.196152422706632)
f(x) = x⁴ – 54x² + 245
Plug in value of x:                   (5.196152422706632)⁴ - 54(5.196152422706632)² + 245 = -484
Plug in value of x:                   (-5.196152422706632)⁴ - 54(-5.196152422706632)² + 245 = -484
Local Minima: (-5.196, -484)
Local Minima: (5.196, 484)
We can now plot all the points! 

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