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

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Answer 1 · 2018-07-12T00:34:30

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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: $" "(-1/2, oo)$