Step-by-step explanation:
Domain of a rational function is everywhere except where we set vertical asymptotes. or removable discontinues
Here, we have
First, notice we have x in both the numerator and denomiator so we have a removable discounties at x.
Since, we don't want x to be 0,
We have a removable discontinuity at x=0
Now, we have
We don't want the denomiator be zero because we can't divide by zero.
so
So our domain is
All Real Numbers except-2 and 0.
The vertical asymptors is x=-2.
To find the horinzontal asymptote, notice how the numerator and denomator have the same degree. So this mean we will have a horinzontal asymptoe of
The leading coeffixent of the numerator/ the leading coefficent of the denomiator.
So that becomes
So we have a horinzontal asymptofe of 2
One A
y = e^x
dy/dx = e^x The f(x) = the differentiated function. Any value that e^x can have, the derivative has the same value. x is contained in all the reals.
One B
y = x*e^x
y' = e^x + xe^x Using the multiplication rule.
You want the slope and the value of the of y to be the same. The slope is y' of the tangent line
xe^x = e^x + xe^x
e^x = 0
This happens only when x is very "small" like x = - 4444444
y = e^x * ln(x) Using the multiplication rule again, we need the slope of the line with is y'
y1 = e^x
y1' = e^x
y2 = ln(x)
y2' = 1/x
y' = e^x*ln(x) + e^x/x So at x = 1 the slope of the line =
y' = e^1*ln(1) + e^1/1
y' = e*0+e = e
y = mx + b
y = ex + b
to find b we use y= e^x ln(x)
e^x ln(x) = e*x + b
e^1 ln(1) = e*1 + b
ln(1) = 0
0 = e + b
b = - e
line equation and answer.
y = e*x - e
The rule for exponents of exponents is:
a^b^c=a^(b*c), in this case:
x^(3/8)^(2/3)
x^((3/8)*(2/3))
x^(6/24)
x^(1/4)
So x^(1/4) is equal to the fourth root.
Answer: I think the answer is the same thing I can’t show a Step-by-step explanation because Math-way requires a premium to show a step-by-step explanation, anyway the answer is 4xy + 2y = 9, I’m sorry about that. :(
Step-by-step explanation: Well anyway, The standard form of a linear equation is Ax+By=C.
