Find the x-values of all points where the curve y=9x^3-9x^2-9x+7  has a horizontal tangent

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txmedteach | High School Teacher | (Level 3) Associate Educator

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This question is equivalent to determining where the slope of the tangent to the curve is 0. The easiest way to do this is through taking the derivative(dy/dx), which gives a function of x whose output is the slope of the line tangent to the curve.

As a crash course, the derivative can be expressed as the following (formally given the function above y(x)):

`dy/dx = lim_(x_(2)->x_(1))(y_(2) - y_(1))/(x_(2) - x_(1))`

If you notice, this looks a lot like the slope function between two points on the curve, but you're seeing what that slope approaches as the second point approaches the first point. The number that the slope approaches is the "derivative" or instantaneous slope at that point.

It is possible to show that the derivative is a "linear operator." This means that given functions f(x) and g(x) and constancs "a" and "b" the following holds true:

`d/dx(af(x) + bg(x)) = ad/dxf(x) + bd/dxg(x)`

In other words, the derivative operator distributes itself pretty much like any multiplicative term! It is also commutative with respect to constants!

For a power of x, we can actually prove that the derivative is the following:

`d/dx x^n = nx^(n-1)`

For example, the derivative of x^2 = 2x.

We combine these two aspects of the derivative to determine the derivative (slope) of the above function (restated here):

`y = 9x^3 - 9x^2 - 9x + 7`

Now, we take the derivative:

`dy/dx = 9*d/dx(x^3) - 9*d/dx(x^2) - 9*d/dx(x) + d/dx(7)`

We now go through and apply the "power of x" rule to each derivative. Notice that when taking the derivative of 7, that's equivalent to taking the derivative of 7*x^0. Based on our formula for the derivative of powers of x, d/dx(7x^0) = 0. Conceptually, this makes sense because a constant function will always have a slope of 0. But we continue by taking the derivative as we have it above:

`dy/dx = 9*(3x^2) - 9*(2x) - 9*(1) + 0`

`dy/dx = 27x^2 - 18x - 9`

Now that we have our derivative, we find where it equals 0 (hence, where the slope of the tangent to the curve is 0):

`0 = 27x^2-18x-9`

Now, we can factor out a 9 and divide both sides of the equation by 9:

`0 = 9(3x^2 - 2x - 1)`

`0 = 3x^2-2x-1`

Now we can factor this trinomial into the product of 2 trinomials:

`0 = (3x+1)(x-1)`

Now you can solve this by recognizing that either 3x+1 or x-1 can be 0 to solve the equation (if a or b = 0, then ab=0). So, we'll solve for both cases:

`3x + 1 = 0`

`3x = -1`

`x = -1/3`

This gives us the solution for the first term. Now, we can move onto the second term:

`x-1 = 0`

`x = 1`

Alternatively, we could have used the quadratic equation if the derivative was not so easily factorable. However, because we were able to factor, we now have both of the x values at which the curve has a horizontal tangent without using that ugly thing:

`x = -1/3`

`x = 1`

Hope that helps!


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