2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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: $$p=0 \text{ and } p=\frac{8}{27}-\frac{4}{9}=-\frac{4}{27}$$

$\endgroup$
1
$\begingroup$

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:

enter image description here

$\endgroup$

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.