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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:)
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