Find the points on the curve
y= 2x^3 + 3x^2 - 36x + 7 where the tangent is horizontal.
dy/dx=6x^2+5x-36=0
solve the quadratic.
To find the points on the curve where the tangent is horizontal, we need to find the points where the derivative of the curve is equal to zero. The derivative gives us the slope of the tangent line at any point on the curve.
Given the equation of the curve: y = 2x^3 + 3x^2 - 36x + 7
Step 1: Find the derivative of the given equation.
The derivative of a function is obtained by differentiating each term with respect to x, using the power rule.
So, taking the derivative of y with respect to x, we get:
dy/dx = d/dx(2x^3) + d/dx(3x^2) - d/dx(36x) + d/dx(7)
= 6x^2 + 6x - 36
Step 2: Set the derivative equal to zero and solve for x.
To find the points where the tangent is horizontal, we need to find the x values where the derivative dy/dx is equal to zero.
So, we can set the derivative equal to zero and solve for x:
6x^2 + 6x - 36 = 0
Step 3: Solve the quadratic equation.
To solve the quadratic equation, we can factor it or use the quadratic formula. However, in this case, it's easier to factor out a common factor of 6:
6(x^2 + x - 6) = 0
(x^2 + x - 6) = 0
Now, we factor the quadratic equation:
(x + 3)(x - 2) = 0
Setting each factor equal to zero:
x + 3 = 0 --> x = -3
x - 2 = 0 --> x = 2
So, we have two possible x-values where the tangent is horizontal: x = -3 and x = 2.
Step 4: Substitute the x-values into the original equation to find the corresponding y-values.
Now, we can substitute these x-values into the original equation y = 2x^3 + 3x^2 - 36x + 7 to find the corresponding y-values.
For x = -3:
y = 2(-3)^3 + 3(-3)^2 - 36(-3) + 7
= -72 + 27 + 108 + 7
= 70
So, when x = -3, y = 70.
For x = 2:
y = 2(2)^3 + 3(2)^2 - 36(2) + 7
= 16 + 12 - 72 + 7
= -37
So, when x = 2, y = -37.
Therefore, the points on the curve where the tangent is horizontal are (-3, 70) and (2, -37).