3x^2 and 3x^3 are both increasing and positive for x>0. How many times greater is the rate of increase of 3x^3 than the rate of increase of 3x^2 at x=4?
y1 = 3x^2 and y2 = 3x^3
at x=4
dy1/dx = 6x or 24
dy2/dx = 9x^2 = 144
so how many times greater ?
144/24 = 6
To find the rate of increase of a function, we can take the derivative of that function with respect to the independent variable. In this case, we need to find the derivatives of the functions 3x^2 and 3x^3, and compare their rates of increase at x=4.
Let's start by finding the derivative of 3x^2. The derivative of x^n with respect to x is given by nx^(n-1). Applying this rule, we get:
d/dx (3x^2) = 2 * 3x^(2-1) = 6x
Next, let's find the derivative of 3x^3:
d/dx (3x^3) = 3 * 3x^(3-1) = 9x^2
Now we have the derivatives of both functions. To compare their rates of increase at x=4, we need to evaluate these derivatives at x=4.
For 3x^2:
f'(4) = 6(4) = 24
For 3x^3:
f'(4) = 9(4)^2 = 144
Now we can compare the rates of increase. The rate of increase of 3x^3 is 144, while the rate of increase of 3x^2 is 24.
To find how many times greater the rate of increase of 3x^3 is than the rate of increase of 3x^2, we can divide the rate of increase of 3x^3 by the rate of increase of 3x^2:
144 / 24 = 6
Therefore, the rate of increase of 3x^3 is 6 times greater than the rate of increase of 3x^2 at x=4.