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.
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.