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I'm fed up with this question from my book. I've calculated the constants to this equation but got stuck at the asymptotes and local extreme values calculations which I need to plot the graph, perhaps anyone could help me out or guide me towards the solution of calculating the asymptotes/local extreme values and then to plot the graph.


Define the constants A,B,C so that a function which is defined by

       (1)  (6/pi) arctan(2-(x+2)²) when x < -1
f(x) = (2)  x + c* |x| - 1          when -1 ≥ x ≥ 1
       (3)  (1/Ax+B) + 4            when x > 1 och Ax + B ≠ 0

is continuous at x = -1 and differentiable in x = 1

I calculated the constants, A,B,C to:

A = -18

B = 16

C = 7/2

Any help is appreciated,

Thanks, Michael.

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Have you tried [WolframAlpha][…? – Berci Nov 22 '12 at 12:59
up vote 1 down vote accepted

Your values for $A,B,C$ are right, at least I got the same values to match the continuity and derivative conditions.

The top arctan function has a single critical point at $x=-2$. This is because the derivative of $\arctan u$ is $1/(1+u^2) \cdot u'$, so the only critical points come from the derivative of the "inside" function, in your case $2-(x+2)^2=-x^2-4x-2.$ This critical point is a local maximum of the top function.

The middle function doesn't have critical points at which derivative is zero, but the value $x=0$ is to be considered a critical point since the derivative of $|x|$ is undefined there. So the graph on the middle part is V shaped, local min at 0.

Neither the top nor the middle have vertical asymptotes, but the bottom function, which by the way has no critical points, has a vertical asymptote where $Ax+B=0$, i.e. at $x=8/9$.

Finally for horizontal asymptotes you can use that arctan itself has asymptote $-\pi/2$ going toward $- \infty$, but the function is multiplied by $6/\pi$ so combine these facts.

And the bottom function, which applies for large positive $x$, has asymptote $y=4$.

EDIT: You should also check for critical points at $x=-1,1$ since it's a piecewise function with functions on pieces changing at these points.

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Note it should also be checked at -1 and 1. I inserted that in the above answer... – coffeemath Nov 22 '12 at 14:53

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