Consider the curve given by the equation x^(3)+xy=-2. It can be shown that (dy)/(dx)=(-3x^(2)-y)/(x). Find the point on the curve where the line tangent to the curve is horizontal.

Set Derivative Equal to Zero: To find where the tangent line to the curve is horizontal, we need to set the derivative dy/dx equal to zero, because a horizontal line has a slope of 0. (dy)/(dx) = (-3x^2 - y) / x = 0Solve for y: Since the numerator of the derivative must be zero for the derivative to be zero, we set the numerator equal to zero and solve for y. -3x^2 - y = 0 y = -3x^2Substitute y back: Now we substitute y = -3x^2 back into the original equation of the curve to find the corresponding x-value. x^3 + x(-3x^2) = -2 x^3 - 3x^3 = -2 -2x^3 = -2Solve for x: Divide both sides by -2 to solve for x^3. x^3 = 1Find y-value: Take the cube root of both sides to solve for x. x = 1Final Point: Now that we have the x-value, we can find the corresponding y-value using the equation y = -3x^2. y = -3(1)^2 y = -3We have found the point on the curve where the tangent line is horizontal. The point is (x, y) = (1, -3).

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Consider the curve given by the equation
x^(3)+xy=-2. It can be shown that
(dy)/(dx)=(-3x^(2)-y)/(x).
Find the point on the curve where the line tangent to the curve is horizontal.

Consider the curve given by the equation $x^{3}+x y=-2$ . It can be shown that $\frac{d y}{d x}=\frac{-3 x^{2}-y}{x}$ . $\newline$ Find the point on the curve where the line tangent to the curve is horizontal. $\newline$ $(\square$ , $\square)$

View step-by-step help

Full solution

Q. Consider the curve given by the equation

x^{3}+x y=-2

. It can be shown that

\frac{d y}{d x}=\frac{-3 x^{2}-y}{x}

\newline

Find the point on the curve where the line tangent to the curve is horizontal.

\newline

(\square

\square)

Set Derivative Equal to Zero: To find where the tangent line to the curve is horizontal, we need to set the derivative $\frac{dy}{dx}$ equal to zero, because a horizontal line has a slope of $0$ . $\frac{dy}{dx} = \frac{-3x^2 - y}{x} = 0$
Solve for y: Since the numerator of the derivative must be zero for the derivative to be zero, we set the numerator equal to zero and solve for y. $\newline$ $-3x^2 - y = 0$ $\newline$ $y = -3x^2$
Substitute $y$ back: Now we substitute $y = -3x^2$ back into the original equation of the curve to find the corresponding $x$ -value. $\newline$ $x^3 + x(-3x^2) = -2$ $\newline$ $x^3 - 3x^3 = -2$ $\newline$ $-2x^3 = -2$
Solve for x: Divide both sides by $-2$ to solve for $x^3$ . $\newline$ $x^3 = 1$
Find y-value: Take the cube root of both sides to solve for $x$ . $x = 1$
Final Point: Now that we have the $x$ -value, we can find the corresponding $y$ -value using the equation $y = -3x^2$ . $\newline$ $y = -3(1)^2$ $\newline$ $y = -3$
Final Point: Now that we have the $x$ -value, we can find the corresponding $y$ -value using the equation $y = -3x^2$ . $\newline$ $y = -3(1)^2$ $\newline$ $y = -3$ We have found the point on the curve where the tangent line is horizontal. The point is $(x, y) = (1, -3)$ .