calculusderivativesgeometrytangent line
Q) find the equations of the tangent lines to the curve $y=x^3+x$ which pass through point $(2,2)$
my attempt: I tried to formulating two equations for both tangents and then inserting the values 2,2 in them.After that I tried to solve these equations but there were too many unknowns and little equations
The derivative of the function is $4x^3 - 12x$, so the slope of the tangent line through the point $(x_0, f(x_0))$ will be exactly $4x_0^3 - 12x_0$. We may actually write the line in slope-intercept form as
$$y - f(x_0) = (4x_0^3 - 12x_0)(x - x_0)$$
or alternatively,
$$y = (4x_0^3 - 12x_0)(x - x_0) + x_0^4 - 6x_0^2$$
So the question is asking under what conditions on $x_0$ the point $(0, 3)$ lies on this curve. Substituting the values, we find that
$$3 = (4x_0^3 - 12x_0)(0 - x_0) + x_0^4 - 6x_0^2$$
Rearranging, this leads to
$$-3x_0^4 +6x_0^2 - 3 = 0$$
Dividing by $-3$ leads to
$$x_0^4 - 2x_0^2 + 1 = 0 \implies (x_0^2 - 1)^2 = 0$$
Hence, we see that $x_0^2 = 1$, so that $x_0 = \pm 1$.
Because sometimes a picture really is worth a thousand words:
Best Answer
Hint:
If those curves are tangent to the curve, their slope is given by $\;y'=3x^2+1\;$ at each general point $\;(x,\,x^3+x)\;$ on the curve , so for what points $\;(a,\,a^3+a)\;$ on the curve are there lines through them and through $\;(2,2)\;$ whose slope is $\;3a^2+1\;$ ?