In Exercises 9–66, graph the function using appropriate method...

Question

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.
y=1-(x+1)^3

y=1-(x+1)^3

Accepted Answer

Graph the function F of X equals 2 plus X minus 2 cubes. The 1st and 2nd derivative are given, where our first derivative is F X equals 3, X minus 2 squad. And F of X is 6 X minus 12. Now to solve this, let's first find a few attributes. We have our domain Which, because this is a polynomial, will be all real numbers, negative infinity to infinity. Because this is a cubic function, our range is also from the infinity to infinity. Hands are symmetry. is none. Now, let's find our critical points and our intervals of increase or decrease. To find a critical point, We will set Our derivative F of X equals zero. So, we have 3 multiplied by X minus 2 squared equals 0. Giving us X minus 22 equals 0, when dividing both sides by 3. And if we take the square root, we get X minus 2 equals 0, or X equals 2. Now, let's check whether it's increasing or decreasing. Since our derivative function is always going to be greater than 0, or equals 0, This means that this Is increasing everywhere. That means that there are no local. Extrema Now, we can look for possible inflection points. To find possible inflection points, we will set our second derivative equals 0.6 X minus 12 equals 0. We can add 12 to both sides, to get 6 X equals 12, and divide by 6 to get X equals 2. Now, let's determine the concavity. If X is less than 2, Our function is concave down. Because our second derivative. Is negative. And the same for X being greater than 2. It will be concave up because the 2nd derivative is greater than 0. No. Let's check if we have asymptotes. Because this is not a rational function, We do not have any asymptotes. And because this is a function of executes. Our end behavior goes from negative infinity to positive infinity. Now let's find our intercepts. For our intercepts we will let X equals 0 for the y intersects. And will let y equals 0 for the X intersect. Let's plug X equals 0 into our function, 2 + x minus 2 cuts. We have 2 + 0 minus 2 cubs, which gives us 2 + -8 or 6, meaning our Y inter set is at 0-6. Our X intercept, we will set 0 equals to our equation. 0 Equals 2 + x minus 2 Q. We can now solve If we solve our function, we get X equals 2 minus the cube root of 2, versus approximately. 0.74, giving us an X intercept of 0.74 0. We now have everything we need to plot this function. Now, we did mention earlier that we had an inflection point at X equals 2. And if we plug that into our equation, We get 2 plus. 2 minus 2 cut, which just equals 2, meaning our specific inflection point is at 22. If we go to a graph, we'll plot the point, add 22. We'll plot another point for our Y intercept at 0-6, and the point for X intercept. At 0.740. We can now draw a curve between our points. This is a cubic function. So we will start from the bottom, go through our points, hit the inflection point, changing cavity, and continue going upwards. OK. I hope to help you solve the problem. Thank you for watching. Goodbye.

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.
y=1-(x+1)^3

Key Concepts

Graphing Functions

Local Extreme Points

Inflection Points

Watch next

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.y=1-(x+1)^3

Key Concepts

Graphing Functions

Local Extreme Points

Inflection Points

Watch next

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.
y=1-(x+1)^3