Question

For what values of x does the curve y^2x^3-15x^2=4 have horizontal tangent lines

Answer 1

We have the following curve:

$y^2x^3-15x^2=4$

To find the tangent lines to the curve we need to use the concept of derivative, but we can’t solve this problem for $y$, thus, let's apply implicit differentiation, so:

$\frac{d}{dx}(y^2x^3-15x^2=4) \\ \\ \frac{d}{dx}(y^2x^3)-\frac{d}{dx}(15x^2)=\frac{d}{dx}(4) \\ \\ (y^{2})'x^3+(x^3)'y^2-15(x^2)'=0 \\ \\ 2yy'x^3+3x^2y^2-30x=0 \\ \\ y'=\frac{dy}{dx}=\frac{30x-3x^2y^2}{2x^3y}=\frac{x(30-3xy^2)}{2x^3y} \\ \\ y'=\frac{30-3xy^2}{2x^2y}$

Therefore the horizontal lines occurs when $y'=0$, then:

$\frac{30-3xy^2}{2x^2y}=0 \\ \\ That \ is, \ when: \\ \\ 30-3xy^2=0 \\ \\ \therefore xy^2=10 \rightarrow y^2=\frac{10}{x}$

If we substitute this in the original equation we have:

$\frac{10}{x}x^3-15x^2=4 \\ \\ \therefore 10x^2-15x^2=4 \\ \\ \therefore x^2=-\frac{4}{5}$

This is an absurd result because it is impossible for a squared number to get a negative number. So the conclusion is that there is no any value of $x$ in which the curve has horizontal tangent lines.