Power Rule for Derivatives. Example Problem 2

Power Rule for Derivatives. Example Problem 2

f(x)=x^3 -9x^2 -48x+5 f'(x)=0 f'(x)=3x^2 -18x-48=0 3(x^2 -6x-16)=0 3(x-8)(x+2)=0 x=8, -2 So. Now you need to find the values of x for which f(x) = x3 – 9×2 – 48x + 5 has a horizontal tangent. If we have a horizontal tangent, that means there’s going to be a slope of 0-that is, f(x) = 0. That means the derivative of f(x) is equal to 0.

We will find f prime of x, differentiate the function with respect to x and set it equal to 0. So the derivative off prime of x equals 3x squared minus 18x minus 48, and we’re going to set that equal to 0. When a number is equal to 0, I have a tendency to want to factor it. Let’s factor the three and the two to get minus 8 and plus 2. The equation does not have a solution when x = 8. However, when x = -2, a solution exists As shown in the graph above, the line has a horizontal tangent when x equals 8 and a slope of 0 when x equals negative 2.