Question

Im having problem to find the second derivative , inflection point, concave up and down intervals.?

Answer 1

See answers below

Explanation:

Given: `f(x) = (x-1)/(2x+1) = (N(x))/(D(x)) = (a_nx^n + ...)/(b_mx^m + ...)`

Analyze the graph analytically:

Find vertical asymptotes `D(x) = 0`:

`2x + 1 = 0; => x = -1/2`

Vertical asymptote at `x = -1/2`

Find horizontal asymptote:

`n = m => " horizontal asymptote ": y = a^n/b^n: " "y = 1/2`

Find the first derivative.
Use the quotient rule `(u/v)' = (vu' - uv')/v^2`

Let `u = x-1; " "u' = 1; " "v = 2x+1; " "v' = 2`

`f'(x) = ((2x+1)(1) - (x-1)(2))/(2x+1)^2`

`f'(x) = (2x + 1 -2x +2)/(2x+1)^2 = 3/(2x+1)^2`

Find the critical values `f'(x) = 0`:

`3/(2x+1)^2 = 0`

`3= 0" "` false

There is no relative maximum or minimum.

Find the second derivative using the power rule:

`f'(x) = 3/(2x+1)^2 = 3(2x+1)^-2`

`f''(x) = -6(2x+1)^-3 (2)`

`f''(x) = -12/(2x+1)^3`

Find inflection points by `f''(x) = 0`

`f''(x) = -12/(2x+1)^3 = 0`

`-12 = 0 " "` false

No inflection points

Find concavity:

This means concavity is only determined by the vertical asymptote

intervals: `(-oo, -1/2), (-1/2, oo)`

test value: `x = -1; " x = 0`

`f''(x) " "+ " " -`

concave up: `" "(-oo, -1/2)`

concave down: `" "(-1/2, oo)`