Answer:
Step-by-step explanation:
When a question asks for the "end behavior" of a function, they just want to know what happens if you trace the direction the function heads in for super low and super high values of x. In other words, they want to know what the graph is looking like as x heads for both positive and negative infinity. This might be sort of hard to visualize, so if you have a graphing utility, use it to double check yourself, but even without a graph, we can answer this question. For any function involving x^3, we know that the "parent graph" looks like the attached image. This is the "basic" look of any x^3 function; however, certain things can change the end behavior. You'll notice that in the attached graph, as x gets really really small, the function goes to negative infinity. As x gets very very big, the function goes to positive infinity.
Now, taking a look at your function, 2x^3 - x, things might change a little. Some things that change the end behavior of a graph include a negative coefficient for x^3, such as -x^3 or -5x^3. This would flip the graph over the y-axis, which would make the end behavior "swap", basically. Your function doesn't have a negative coefficient in front of x^3, so we're okay on that front, and it turns out your function has the same end behavior as the parent function, since no kind of reflection is occurring. I attached the graph of your function as well so you can see it, but what this means is that as x approaches infinity, or as x gets very big, your function also goes to infinity, and as x approaches negative infinity, or as x gets very small, your function goes to negative infinity.
306+122=428
Because you add 6 & 2 in the ones place which is 8.
then 2 and 0 which is 2.
finally 1 and 3 which is 4
Answer:
Step-by-step explanation:
I see you're in college math, so we'll solve this with calculus, since it's the easiest way anyway.
The position equation is
That equation will give us the height of the rock at ANY TIME during its travels. I could find the height at 2 seconds by plugging in a 2 for t; I could find the height at 12 seconds by plugging in a 12 for t, etc.
The first derivative of position is velocity:
v(t) = -3.72t + 15 and you stated that the rock will be at its max height when the velocity is 0, so we plug in a 0 for v(t):
0 = -3.72t + 15 and solve for t:\
-15 = -3.72t so
t = 4.03 seconds. This is how long it takes to get to its max height. Knowing that, we can plug 4.03 seconds into the position equation to find the height at 4.03 seconds:
s(4.03) = -1.86(4.03)² + 15(4.03) so
s(4.03) = 30.2 meters.
Calculus is amazing. Much easier than most methods to solve problems like this.
Answer:
See below, I will let graphing part to yourself.
Step-by-step explanation:
First function: domain:
, range:
, decreasing
Second function: domain:
, range:
, increasing
Third function: domain:
, range:
, increasing