What are the local extrema of $f(x)= (x^5-x^2-4)/(x^3-3x+4)$ ?

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Answer 1 · 2016-02-10T16:50:06

Local maximum $~~ -0.794$ (at $x~~ -0.563$ ) and local minima are $~~ 18.185$ (at $x~~ -3.107$ ) and $~~ -2.081$ (at $x~~0.887$ )

$f'(x) = (2x^7-12x^5+21x^4+15x^2-8x-12)/(x^3-3x+4)^2$

Critical numbers are solutions to

$2x^7-12x^5+21x^4+15x^2-8x-12 = 0$ .

I do not have exact solutions, but using numerical methods will find real solutions are approximately:

$-3.107$ , $- 0.563$ and $0.887$

$f''(x) = (2x^9-18x^7+14x^6+108x^5-426x^4+376x^3+72x^2+96x-104)/(x^3-3x+4)^3$

Apply the second derivative test:

$f''(-3.107) > 0$ , so $f(-3.107) ~~ 18.185$ is a local minimum

$f''(- 0.563) < 0$ , so $f(- 0.563) ~~ -0.794$ is a local maximum

$f''(0.887) > 0$ , so $f(0.887) ~~ -2.081$ is a local minimum

What are the local extrema of f(x)= (x^5-x^2-4)/(x^3-3x+4)f(x)= (x^5-x^2-4)/(x^3-3x+4)?