2 Answers | Add Yours
You need to remember that if you need to find the vertical asmptotes to the graph of rational function, you should look for the roots of denominator such that:
`x^2 + 2x = 0`
You need to factor out x such that:
`x(x+2) = 0`
`x = 0 or x + 2 = 0 =gt x = -2`
Hence, the vertical lines x = 0 and x = -2 are the vertical asymptotes of function.
b) You need to remember that the graph of function is concave down for f''(x)<0 such that:
`f'(x) = ((x^3+3x^2+3x+1)'*(x^2+2x) - (x^3+3x^2+3x+1)*(x^2+2x)')/((x^2+2x)^2)`
`f'(x) = ((3x^2 + 6x + 3)*(x^2+2x) - (x^3+3x^2+3x+1)*(2x+2))/((x^2+2x)^2)`
`f'(x) = (3x^4 + 6x^3 + 6x^3 + 12x^2 + 3x^2 + 6x - 2x^4 - 2x^3 - 6x^3 - 6x^2 - 6x^2 - 6x - 2x - 2)/((x^2+2x)^2)`
`f'(x) = (x^4 + 4x^3 + 3x^2 - 2x - 2)/((x^2+2x)^2)`
You need to evaluate the second derivative such that:
`f''(x) = ((4x^3 + 12x^2 + 6x - 2)(x^2+2x)^2 - 4(x^2+2x)(x^4 + 4x^3 + 3x^2 - 2x - 2)(x+1))/((x^2+2x)^4)`
`f''(x) = ((4x^3 + 12x^2 + 6x - 2)(x^2+2x) - 4(x^4 + 4x^3 + 3x^2 - 2x - 2)(x+1))/((x^2+2x)^3)`
`4x^5 + 8x^4 + 12x^4 + 24x^3 + 6x^3 + 12x^2 - 2x^2 - 4x - 4x^5 - 4x^4 - 16x^4 - 16x^3 -12x^3 - 12x^2 + 8x^2 + 8x + 8x + 8 = 0`
`24x^3 + 6x^3 + 12x^2 - 2x^2 - 4x - 16x^3 -12x^3 - 12x^2 + 8x^2 + 8x + 8x + 8 = 0`
`-4x^3 + 6x^2 + 12x + 8 = 0`
`2x^3 - 3x^2 - 6x - 4 = 0`
If `x in [-16,17] =gt f''(x)lt0` , hence the graph of the function is concave down.
- For function to have a vertical asymptote, it needs to diverge into either when x approaches certain value. And for the function in fractional form, this can be reached by making the denominator 0 while keeping numerator with some value other than 0.
=> x=0 or -2
With each value, numerator doesn't become 0.
Therefore, f(x) has its vertical asymptote at x=0 and x=-2
lim(x->0)f(x) = diverge in positive way,
lim(x->-2)f(x)=diverge in negative way.
- f(x) is concave down when f''(x), the second derivative of f(x), has negative value.
Therefore, f(x) is concave down at -1<x<0 or -16<x<-2.
Hope you do great on your work:)
We’ve answered 317,763 questions. We can answer yours, too.Ask a question