Increasing/Decreasing/Relative Extrema/Convavity/Inflection Points/Additional PointsIncreasing/Decreasing/Relative Extrema/Convavity/Inflection Points/Additional Points for these two problems:...

Increasing/Decreasing/Relative Extrema/Convavity/Inflection Points/Additional Points

Increasing/Decreasing/Relative Extrema/Convavity/Inflection Points/Additional Points

for these two problems:

f(x): (2x^2-5x+5)/(x-2)

and

f(x): (x)+(32/x^2)

 

I understand how to do curve sketching, these are just giving a heck of a time. Thank you in advance for your answers:)

Asked on by lindsey0

1 Answer | Add Yours

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

1)

y = (2x^2-5x+5)/(x-2).

= (2x-1) + 3/(x-2). Differentiating with respect to x, we get:

y' = 0 gives 2 +3*(-1)/(x-2)^2 = 0.  Or 2(x-2)^2 = 3 Or x = 2+(3/2)^(1/2). Or x = 2-(3/2) ^(1/2).

y'' = d/dx{2 - 3/(x-2)^2} =  0 - 3(-2)/(x-2)^3 = 6/(x-2)^3 is  clearly positive for x= 2+(3/2)^(1/2) and  negative for x = 2-(3/2)^(1/2).

Case (i) x > 2

So  when x = 2+(3/2)^(1/2),  y = 2x-1 + 3/(x-2)  =   2(2+(3/2)^(1/2)) -1 +3/(3/2)^(1/2) = 4+6^(1/2) -1 + 6^(1/2) = 3+2*6(1/2) is the minimum.

The curve y = 2x-1 +3/(x-2) undefined at x=2. But from x=  2+ to 2+(3/2)^(1/2) y value is decreasing from unbounded height to a minimum  3+2*6^(1/2) and again  increasing and approaches asymptotically to the line 2x-1. Thus the curve is enclosed between the vertical asymptote x=2 and an oblique asymptote 2x-1.

As y'' > 0, the concavity is up.

case (ii): x < 2

Similarly when x = 2-(3/2)^(1/2), the curve y = 2x-1+3/(x-2) attains its maximum  when x = 2-(3/2)^(1/2), as y'' = 6/(x-2)^3 is negative for this value of x.

So the maximum y value = 2[2-(3/2)^(1/2)] -1 +3/(-(3/2)^(1/2)) = 4-sqrt6-1 - sqrt6 = 3 - 2sqrt6.

At x = 2 the curve y = 2x-1+2/(x-2)  does not exist.

As x approaches from 2- towards 2-(3/2)^(1/2) the curve increases from negative infinity to a maximum of 3-2*(6)^(1/2). And then from x= 2-(3/2)^(1/2) to  minus infinity, the curve decreases and approaches  asymptotically 2x-1.

When x < 2, the curve remains enclosed between the oblique asymptote 2x-1 and vertical asymptote x=2 with its concavity down ward as y'' < 0.

2)

y = x+32/x^2. y' = 0 gives d/dx(x+32/x^2) = 0. Or

1+32(-2)/x^3 = 0 Or x^3 = 64. So x = 4.

At x=4, y" =[ d/dx(-64/x^3) at x=4 ] = 192/4^4 = 3/4 a positive.

So at x = 4, y =x+32/x^2 = 4+32/4^2 = 6 is the minimum.

At x = 0, y does not exist.

Case (i) x>0.

From 0+ to x=4, th e curve is decreasing from positive infinity to the minimum  6 and from x=4 onwards, as x approaches infinity y also increases and approaches asymptotically y = x.

Thus the curve is enclosed by x=0, the vertical asymotote y axis and the oblique asymptote y = x line, Since the y ''  is positive , when x> 0, the concavity of the curve is upward.

Case (ii) x< 0.

y = x+32/x^2 is positively unbounded at 0-

When y=0=x+32/x^2. Or x = -32^(1/3) = 3.1748 nearly. So the curve intercepts x axis at -32^(1/3) .

y' = 1-64/x^3  positive for all x <0. So the curve is an increasing function and so decreases as x decreases or increases as x increases. y = x+32/x^2 ultimately approaches the oblique asymptote y = x as x approaches minus infinity.

Since y" > 0 the concavity of the curve is up.

We’ve answered 318,957 questions. We can answer yours, too.

Ask a question