Question
Find a constant $b$ so that $\sin(3x)$ and $1 - e^{bx}$ overlap on the interval $(-1,1)$ when graphed.
So I know that for the graphs to overlap the derivatives need to be equal. Not sure how to start this. I basically just picked a few random b's and guessed through trial and error, but no luck.
You can show that both curves $f(x)=\sin(3x)$ and $g(x)=1-e^{bx}$ pass through the origin. So we know they cross each other at least at the origin $(0,0)$. At this intersection point, for them to overlap each other, their gradients must be the same. The gradients of these functions are given by:$$\frac{df}{dx}=3\cos(3x)\tag{1}$$$$\frac{dg}{dx}=-be^{bx}\tag{2}$$These gradients must be equal for the curves to be considered as overlapping, therefore:$$3\cos(3x)=-be^{bx}$$And, for them to overlap at the origin we must have:$$\begin{align} 3\cos(0)&=-be^0\\ \therefore 3&=-b\\ \therefore b&=-3 \end{align}$$ There may be other points at which these curves overlap as I have only considered the origin above.
Following on from my analysis above, we know at the points of overlap:$$f(x)=g(x)$$$$\frac{df}{dx}=\frac{dg}{dx}$$This leads to:$$\sin(3x)=1-e^{bx}$$$$3\cos(3x)=-be^{bx}=-b(1-\sin(3x))\text{ (from equation above)}$$Square both sides to get:$$9\cos^2(3x)=b^2(1-2\sin(3x)+\sin^2(3x))$$which reduces down to this quadratic in $\sin(3x)$:$$(9+b^2)\sin^2(3x)-2b^2\sin(3x)-(9-b^2)=0$$I used the standard equation to solve this quadratic to get:$$\sin(3x)=\frac{b^2\pm9}{b^2+9}$$which leads to either:$$\sin(3x)=1\tag{3}$$or:$$\sin(3x)=\frac{b^2-9}{b^2+9}\tag{4}$$Solution (3) leads to $e^{bx}=0\implies b=-\infty$ and I have not found any easy way to solve for $b$ using solution (4).