Given the equation $$2y^3+y^2-y^5=x^4-2x^3+x^2$$ I have to find the x-values for which the slope of the tangent line is equal to 0 (horizontal).
I derived the equation… $$(4x^3-6x^2+2x)\over(6y^2+2y-5y^4)$$
…but while with an equation with one variable, I would set the equation equal to 0 and solve for x, I don't know what to do with two variables.
[Math] How to find horizontal tangent line with two variables
calculusderivatives
Related Solutions
Simplify $f'(x)=(x-2)(2x-1)+(1)(x^2-x-11)$
- $f'(x) = (2x^2 -5x + 2) + (x^2 - x - 11)$
- $f'(x) = 3x^2 - 6x - 9$
And find where $f'(x) = 0$
$3x^2 - 6x - 9 = 0 \iff x^2 - 2x - 3 = (x + 1)(x - 3) = 0$
That's where slope is 0, hence any line tangent at that point will be horizontal: when $x = 3$ or when $x = -1$.
So the roots (x values) of the points you need are
$x_1 = 3$, and
$x_2 = -1$.
Then find the corresponding $y$ value in the ORIGINAL equation: $$\;f(x)=(x-2)(x^2-x-11),\;$$ to determine the two points: $(x_1, y_1), (x_2, y_2)\;$ where the line tangent to $f(x)$ is horizontal.
$y_1 = f(3) = 1 \cdot -5 = -5$
$y_2 = f(-1) = (-3)(-9) = 27$
So your points are $(3, -5)$ and $(-1, 27)$.
I've included a graph of the function $\;f(x)=(x-2)(x^2-x-11)\;$(in blue), along with the two horizontal lines tangent to the function: $\;y = -5,\;\; y = 27$ (violet and "brown", respectively) to see in a "picture" what is happening here.
The slope of a line perpendicular to $y = mx + b$ is equal to $-\frac 1m$, where slope $m$ of the equation $y = 3x - 1$ is $m = 3$.
So solve for the points at which the first derivative is equal to $-\dfrac 1m = -\dfrac 13$. You'll get a quadratic equation, from which you can find the two solutions.
$$y'=\frac{x-4}{(x-3)^2} = -\dfrac 13$$
$$ (x - 3)^2 = -3( x- 4),\;\;x \neq 3$$ $$x^2 - 6x + 9 = -3x + 12$$ $$ \iff x^2 - 3x - 3 = 0$$
Those solutions, (solving for $x$), will give you the $x$-coordinate of the points at which the tangent lines to the curve are perpendicular to the given line. We can simply use the quadratic equation to find the solutions, $x_1$ and $x_2$.
To find the corresponding $y$ coordinate, just plug each solution $x_i$ into the original curve (not into $y'$) and compute $y$.
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Best Answer
Note that if $$y'=\frac{(4x^3-6x^2+2x)}{(6y^2+2y-5y^4)}$$ then $$y'=0$$ if and only if
$$4x^3-6x^2+2x=0$$