Consider the curve given by the equation 2x^(2)-12 x+y^(3)=107. It can be shown that (dy)/(dx)=(4(3-x))/(3y^(2)). Find the point on the curve where the line tangent to the curve is horizontal.

Solve for x: Solve for x: 4(3 - x) = 0 12 - 4x = 0 4x = 12 x = 3Find y-coordinate: Now that we have the x-coordinate, we need to find the corresponding y-coordinate on the curve. We plug x = 3 into the original equation of the curve to find y. 2x^(2) - 12x + y^(3) = 107 2(3)^(2) - 12(3) + y^(3) = 107Calculate y^(3): Calculate the value of y^(3): 2(9) - 36 + y^(3) = 107 18 - 36 + y^(3) = 107 y^(3) = 107 + 36 - 18 y^(3) = 125Find y: Find the cube root of 125 to get y: y = 125^(1/3) y = 5

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Consider the curve given by the equation
2x^(2)-12 x+y^(3)=107. It can be shown that
(dy)/(dx)=(4(3-x))/(3y^(2)).
Find the point on the curve where the line tangent to the curve is horizontal.

Consider the curve given by the equation $2 x^{2}-12 x+y^{3}=107$ . It can be shown that $\frac{d y}{d x}=\frac{4(3-x)}{3 y^{2}}$ . $\newline$ Find the point on the curve where the line tangent to the curve is horizontal. $\newline$ $(\square$ , $\square)$

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Calculus

Find indefinite integrals using the substitution

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Q. Consider the curve given by the equation

2 x^{2}-12 x+y^{3}=107

. It can be shown that

\frac{d y}{d x}=\frac{4(3-x)}{3 y^{2}}

\newline

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

\newline

(\square

\square)

Solve for x: Solve for x: $\newline$ $4(3 - x) = 0$ $\newline$ $12 - 4x = 0$ $\newline$ $4x = 12$ $\newline$ $x = 3$
Find y-coordinate: Now that we have the x-coordinate, we need to find the corresponding y-coordinate on the curve. We plug $x = 3$ into the original equation of the curve to find $y$ . $\newline$ $2x^{2} - 12x + y^{3} = 107$ $\newline$ $2(3)^{2} - 12(3) + y^{3} = 107$
Calculate $y^{3}$ : Calculate the value of $y^{3}$ : $2(9) - 36 + y^{3} = 107$ $18 - 36 + y^{3} = 107$ $y^{3} = 107 + 36 - 18$ $y^{3} = 125$
Find $y$ : Find the cube root of $125$ to get $y$ : $y = 125^{(1/3)}$ $y = 5$