Question

Find the points on the curve \(y=2 x^{3}-3 x^{2}-12 x+20\) where the tangent is parallel to the \(x\) -axis.

Short answer

The points on the curve where the tangent is parallel to the x-axis are \((2, 4)\) and \((-1, 21)\).

Step-by-step solution

Find the derivative of the function

The derivative of \(y = 2x^3 - 3x^2 - 12x + 20\) is found using power rules of differentiation. The derivative, \(y'\), is \(y' = 6x^2 - 6x - 12\).

Set the derivative equal to 0

To find where the tangent is parallel to the x-axis(i.e., slope is 0), we set \(y' = 0\), which gives \(6x^2 - 6x -12 = 0\). As you can see, we can factor out a 6, simplifying it to \(x^2 - x - 2 = 0\).

Solve the equation for x

We now have a simple quadratic equation which we can solve using factoring, completing the square or quadratic formula. Using the factoring method, we get \((x -2)(x +1) = 0\), hence \(x = 2, -1\).

Substitute x values back into the original function

Substitute \(x = 2, -1\) back into the original equation to get the corresponding y values. For \(x = 2, y = 2(2^3) - 3(2^2) - 12*2 + 20 = 4\). For \(x = -1, y = 2(-1)^3 - 3(-1)^2 - 12*-1 + 20 = 21\).

State the final result

Therefore, the points on the curve \(y = 2x^3 - 3x^2 - 12x + 20\), where the tangent is parallel to the x-axis are \((2,4)\) and \((-1,21)\).