0
$\begingroup$

The question is to "Find the values of k for which the line $y=3x$ is tangent to the cubic $y=x^3+k$". By differentiating (giving $\frac{dy}{dx}=3x^2$) I can work out that $\frac{dy}{dx}=3$ at 1 and -1... but I can't see how to work out values of k from this information. Any hints?

$\endgroup$
  • $\begingroup$ Well, what are the equations of the tangent lines at these points? $\endgroup$ – amd Oct 3 '17 at 20:25
0
$\begingroup$

the slope of your function $$y=x^3+k$$ must be equal to the slope of the given Tangent line, this means $$3x^2=3$$

$\endgroup$
  • $\begingroup$ The OP has already found the values of $x$ at which the slope is $3$. The question is about how to adjust the constant term of the cubic so that the tangent there passes through the origin. $\endgroup$ – amd Oct 3 '17 at 20:24
0
$\begingroup$

Use the fact that the tangent line touches the curve at 2 points, in this case, ($1,3$) and ($-1,-3$). Can you proceed now?

$\endgroup$

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.