Question is Find the $x$-coordinate of all points on the curve $$y=22x\sin(5x)+55\sqrt{3}x^2+68,\quad \frac\pi{10}<x<\frac{3\pi}{10}$$ where the tangent line passes through the point $P(0,68)$ (not on the curve) there are two $x$ values what are they?
So the point that's really confusing is me is I got the derivative of the function but how do I get the points when only given a point NOT on the curve??
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$\begingroup$Outline: (1) For brevity we let $y=f(x)$. You have found the slope $f'(x)$ of the tangent line at the point $(x,y)$ on the curve with $x$-coordinate equal to $x$.
(2) You know that the tangent line passes through $(0,68)$.
(3) It follows that $$\frac{f(x)-68}{x-0}=f'(x).$$
(4) Solve for $x$. We get "lucky," since $\frac{f(x)-68}{x}$ simplifies nicely.
$\endgroup$ 3 $\begingroup$From the equation you give as $y=f(x)$, first write an equation for the tangent line to the graph at the point $(a,f(a))$ in the form $y=f(a)+f'(a)(x-a)$, and note that the point $P(0,68)$ (not on the graph, as you noted) is supposed to be on that line, so $68=f(a)+f'(a)(0-a)$ -- now solve for $a$.
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