# 1. What is 1.0886621 In fraction form. The answer is wanted in a form such as 16`sqrt(x)` /9 and I don't know how to find it quickly or find it at all. 2. Is the Point of Inflection for 3x/(x^2-9)...

1. What is 1.0886621 In fraction form. The answer is wanted in a form such as 16`sqrt(x)` /9 and I don't know how to find it quickly or find it at all.

2. Is the Point of Inflection for 3x/(x^2-9) coordinates (0,0)?

embizze | Certified Educator

(a) Express the approximation 1.0886621 in the form `(16sqrt(x))/9 ` :

( I assume that you have done a problem and used a calculator to find an approximate answer, while the text or computer program is expecting an exact answer. You will want to go back and see if you can arrive at the exact answer using algebraic techniques. Your instructor may accept approximate answers, but some computer homework programs will not.)

`1.0886621=(16sqrt(x))/9 `  Multiply both sides by 9:

` 16sqrt(x)=9.7979589 ` Divide both sides by 16:

`sqrt(x)=.6123724313 ` Square both sides:

`x=.3749999946~~.375=3/8 `

So `1.0886621~~(16sqrt(3/8))/9=4sqrt(6)/9 `

This is a useless technique to learn, unless you are always given the final form of the answer.

(b) Find any inflection points for the graph of `y=(3x)/(x^2-9) ` :

Inflection points are points where the graph changes concavity. They can be found algebraically by setting the second derivative equal to zero. If the second derivative changes sign at such a point, it is an inflection point. (These are points where the rate of rate of change is changing.)

`y'=((x^2-9)(3)-(3x)(2x))/((x^2-9)^2)=(-3(x^2+9))/((x^2-9)^2) ` (Using the quotient rule.)

`y''=(6x(x^2+27))/((x^2-9)^3) `

Setting the second derivative equal to zero:

Note that for fractions, the fraction is zero if the numerator is zero while the denominator is nonzero.

`6x(x^2+27)=0 ==> x=0 `

For -3<x<0, the second derivative is positive (implying the graph is concave up.)

For 3>x>0 the second derivative is negative (implying the graph is concave down.)

Since the second derivative is zero at x=0, and the second derivative changes sign at x=0, there is an inflection point at x=0.

On the graph there are vertical asymptotes at x=-3 and x=3; the graph is concave down for x<-3, concave up for -3<x<0, concave down for 0<x<3, and concave up for x>3.

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There is an inflection point at (0,0)

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The graph: