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DochEvi [55]
3 years ago
5

What are the vertical asymptotes of the function f(x) =5x+5/x2 + x-2

Mathematics
1 answer:
katen-ka-za [31]3 years ago
5 0

let's recall that the vertical asymptotes for a rational expression occur when the denominator is at 0, so let's zero out this one and check.

\bf \cfrac{5x+5}{x^2+x-2}\qquad \stackrel{\textit{zeroing out the denominator}~\hfill }{x^2+x-2=0\implies (x+2)(x-1)=0}\implies \stackrel{\textit{vertical asymptotes}}{ \begin{cases} x=-2\\ x=1 \end{cases}}

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Answer:

$2.1825 or simplified is $2.18.

Step-by-step explanation:

<em>0.7275 per candy bars.</em>

<em>0.7275  x 4 = $2.91.</em>

So according to that logic,

<em>0.7275  x 3 = $2.1825.</em>

<em />

6 0
3 years ago
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Read 2 more answers
Plz help!
beks73 [17]

First, you need to find the derivative of this function.  This is done by multiplying the exponent of the variable by the coefficient, and then reducing the exponent by 1.  

f'(x)=3x^2-3

Now, set this function equal to 0 to find x-values of the relative max and min.

0=3x^2-3

0=3(x^2-1)

0=3(x+1)(x-1)

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To determine which is the max and which is the min, plug in values to f'(x) that are greater than and less than each.  We will use -2, 0, 2.

f'(-2)=3(-2)^2-3=3(4)-3=12-3=9

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f'(2)=3(2)^2=3(4)-3=12-3=9

We examine the sign changes to determine whether it is a max or a min.  If the sign goes from + to -, then it is a maximum.  If it goes from - to +, it is a minimum.  Therefore, x=-1 is a relative maximum and x=1 is a relative miminum.

To determine the values of the relative max and min, plug in the x-values to f(x).

f(-1)=(-1)^3-3(-1)+1=-1+3+1=3

f(1)=(1)^3-3(1)+1=1-3+1=-1

Hope this helps!!

7 0
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tekilochka [14]

Answer:

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Step-by-step explanation:

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