Questions: Find the points on the curve (y=2 x^3+3 x^2-12 x+5) where the tangent line is horizontal. ((x, y)=()( smaller (x) -value) ) ((x, y)=()( larger (x) -value) )

Solution Steps

To find the points on the curve where the tangent line is horizontal, we need to find the points where the derivative of the function is zero. The derivative of the function gives us the slope of the tangent line, and a horizontal tangent line has a slope of zero.

1. Compute the derivative of the function \( y = 2x^3 + 3x^2 - 12x + 5 \).
2. Set the derivative equal to zero and solve for \( x \).
3. Substitute the \( x \)-values back into the original function to find the corresponding \( y \)-values.

Given the function \( y = 2x^3 + 3x^2 - 12x + 5 \), we first compute its derivative: \[ \frac{dy}{dx} = 6x^2 + 6x - 12 \]

To find the points where the tangent line is horizontal, we set the derivative equal to zero and solve for \( x \): \[ 6x^2 + 6x - 12 = 0 \] Solving this quadratic equation, we get: \[ x = -2 \quad \text{and} \quad x = 1 \]

Next, we substitute these \( x \)-values back into the original function to find the corresponding \( y \)-values.

For \( x = -2 \): \[ y = 2(-2)^3 + 3(-2)^2 - 12(-2) + 5 = 2(-8) + 3(4) + 24 + 5 = -16 + 12 + 24 + 5 = 25 \]

For \( x = 1 \): \[ y = 2(1)^3 + 3(1)^2 - 12(1) + 5 = 2(1) + 3(1) - 12 + 5 = 2 + 3 - 12 + 5 = -2 \]

Final Answer