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I have a doubt on the following problem:

Given $f(x) = x^4 - 6x^2$, find the tangents which pass through point $(0, 3)$.

I'm only used to find tangent equations for points inside the curve, how to do when they are outside ?

Thanks in advance.

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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 $x_0 = \pm 1$.

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