# Calculate asymptotes and local extreme values

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.

Equation:

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][wolframalpha.com/input/…? – Berci Nov 22 '12 at 12:59

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