For which values of $p$ is the line $y_p(x)$ tangent to $f(x)$

Question

I tried thinking about this problem and I just don't get it.

For which values of $p$ is the line $y=px$ tangent to the graph of $f(x)$?

$f(x)=x^4-x^3$

Okay so I usually, when I get a question like "for which values are these two functions tangent", I take the derivative of both functions, equal them to each other and see if both functions have the same $y$ value at that point.

So let's do that:

$f'(x)=y_p'(x)=4x^3-3x^2=p$

Now I'm kind of stuck? Okay so the Slope is equal to each other when $p = 4x^3-3x^2$. I really don‘t get how to continue.

plugging in the $p$ into $y=px$ gives you the results in terms of $x$ to which you get some values. What do I do with these values and why? I’m pretty confused about all of this.

thanks in advance for help.

Answer 1

You need more than just the slopes to be equal. At these points, y must also be equal. So you also have:

$$px=x^4-x^3$$ $$p=x^3-x^2$$

This, with your equation, is a system of 2 equations and 2 unknowns, $p$, and $x$. It is set up already for substitution for $p$ so:

$$x^3-x^2= 4x^3-3x^2$$ $$3x^3-2x^2=0$$ $$x^2(3x-2)=0$$ $$x=0 \text{ and } x=\frac{2}{3}$$ These are your two points where the slopes are the same and the $y$-values are the same. Easy to get the two $p$-values by substitution into above: $$p=0 \text{ and } p=\frac{8}{27}-\frac{4}{9}=-\frac{4}{27}$$

Answer 2

Guide:

At point $t$, the tangent is

$$y-(t^4-t^3)=(4t^3-3t^2)(x-t)$$

$$y=(4t^3-3t^2)x+(t^4-t^3-4t^4+3t^3)$$

$$y=(4t^3-3t^2)x+(-3t^4+2t^3)$$

Let $-3t^4+2t^3=0$ to solve for $t$.

Cool picture: