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xy = 32 makes y = 32/x
z² = (32/x)² + (x/2)²
<span>Isolate z
</span>z = 32/x + x/2
Find dz/dx
fr(z) = (x² - 64)/ (2x²)
<span>Set z' = 0, algebraically solving for x having
</span>x +- 18
<span>Since you cannot have a dimension of -8, the only answer for x is 8.
</span>
<span>Go back to the original area equation; 32 = xy and plug in x making
</span><span>32 = 8y
</span><span>y = 4</span>
Answer:
At x = 4 ⇒ g(x) = f(2)
Step-by-step explanation:
* In the functions f(x) = 3x - 1 and g(x) = 2x - 3, the domain is
the values of x and the range is the values of y
∵ g(x) = f(2)
* Then you will use the range of f(x) at x = 2 to find the domain of g(x)
∵ The domain of f(x) = 2
∵ f(x) = 3x - 1
∴ f(2) = 3(2) - 1 = 6 - 1 = 5
* The range of f(x) = 5
∵ f(2) = g(x)
∴ 5 = g(x)
∵ g(x) = 2x - 3
∴ 2x - 3 = 5 ⇒ add 3 to both sides
∴ 2x = 8 ⇒ divide both sides by 2
∴ x = 4
* At x = 4 ⇒ g(x) = f(2)
Answer:
Increasing: ![(\infty, -1) \cup (2, \infty)](https://tex.z-dn.net/?f=%28%5Cinfty%2C%20-1%29%20%5Ccup%20%282%2C%20%5Cinfty%29)
Decreasing: ![(-1, 2)](https://tex.z-dn.net/?f=%28-1%2C%202%29)
Step-by-step explanation:
So when an equation has and odd degree, it will go in the opposite direction on both ends, so if y went towards infinity as x went towards infinity, then y would go towards negative infinity as x goes towards negative infinity. In this case, by looking at the graph it has an odd degree, due to opposite end behaviors, although on both ends it's increasing because even though it appears that it's going down on the left side, that's only if you start from the right and go towards the left. So it's really increasing from negative infinity to -1, and then it decreases from -1 to 2, until it once again starts increasing from 2 to infinity. This can be represented as (-infinity, -1) U (2, infinity) for increasing and (-1, 2) as decreasing