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.


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 $x_0 = \pm 1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.