Answer:
-4
Step-by-step explanation:
i just took this
<h3>
Answer: Choice C</h3>
Work Shown:
So that's why the answer is choice C
The requirement that x is nonzero isn't technically necessary. The original expression simplifies to choice C even when x = 0 is the case. Also, we don't have issues such as division by zero errors that could arise. It's a bit curious why your teacher put in that condition.
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
Are you sure this is even possibly
