Answer :
Final Answer:
The rate of change of [tex]\(f'(x)\)[/tex] at [tex]\(x = 2\)[/tex]is [tex]\(64\)[/tex], calculated from the derivative[tex]\(f'(x) = 38x - 3x^2\).[/tex]
Explanation:
To find the rate of change of [tex]\(f'(x)\)[/tex]at [tex]\(x = 2\)[/tex], we first need to find [tex]\(f'(x)\),[/tex] which represents the derivative of[tex]\(f(x) = 19x^2 - x^3\)[/tex].
First, find [tex]\(f'(x)\)[/tex] by differentiating[tex]\(f(x)\)[/tex] with respect to[tex]\(x\)[/tex]:
[tex]\[f'(x) = \frac{d}{dx}(19x^2 - x^3)\][/tex]
[tex]\[f'(x) = 38x - 3x^2\][/tex]
Now, evaluate[tex]\(f'(x)\)[/tex] at [tex]\(x = 2\)[/tex]:
[tex]\[f'(2) = 38(2) - 3(2^2)\][/tex]
[tex]\[f'(2) = 76 - 12\][/tex]
[tex]\[f'(2) = 64\][/tex]
So, the rate of change of[tex]\(f'(x)\)[/tex] at [tex]\(x = 2\) is \(64\).[/tex]
Therefore, the rate of change of [tex]\(f'(x)\)[/tex] at[tex]\(x = 2\) is \(64\)[/tex]. This means that at the point[tex]\((2, 68)\)[/tex] on the graph of [tex]\(f(x) = 19x^2 - x^3\),[/tex] the slope of the tangent line (which represents the rate of change) is [tex]\(64\)[/tex]. This indicates how quickly the function [tex]\(f(x)\)[/tex] is changing at the specific point [tex]\(x = 2\)[/tex], emphasizing its steep incline at that location.
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Final Answer:
The rate of change of f'(x) at (2, 68) is -8. This indicates a decreasing slope for the function f'(x) at x = 2.
Explanation:
In calculus, the rate of change of a function can be determined by finding the derivative of that function and evaluating it at the specific point of interest. Given the function f(x) = 19[tex]x^2[/tex] - [tex]x^3[/tex], we first find its derivative f'(x). The derivative of f(x) with respect to x is f'(x) = 38x - 3[tex]x^2[/tex].
To find the rate of change of f'(x) at the point (2, 68), we substitute x = 2 into the expression for f'(x), which gives us f'(2) = 38(2) - 3[tex](2)^2[/tex] = 76 - 12 = 64. Thus, the rate of change of f'(x) at x = 2 is 64.
However, the question asks for the rate of change of f'(x) at (2, 68), meaning we need to evaluate f'(x) at x = 2 and then determine the corresponding y-value. Plugging x = 2 into the original function f(x), we find that f(2) = 19[tex](2)^2[/tex] - [tex](2)^3[/tex] = 76 - 8 = 68.
Therefore, at the point (2, 68), the rate of change of f'(x) is the slope of the tangent line to the graph of f'(x) at x = 2, which is the value we obtained earlier, 64. However, we should note that the rate of change is negative (-64) because f'(x) is decreasing at x = 2. Thus, the rate of change of f'(x) at (2, 68) is -64.
