Respuesta :
Answer: [tex]0.024\ \text{miles per sec}[/tex]
Step-by-step explanation:
Given
Path is changing according to the function [tex]f(x)=0.03(10x-x^2)[/tex]
Rate of change of the elevation is given by the derivative of the function
[tex]\Rightarrow f'(x)=0.03(10-2x)[/tex]
At [tex]x=1\ f'(x) \ \text{is}[/tex]
[tex]\Rightarrow f'(x=1)=0.03(10-2\times 1)\\\Rightarrow f'(x=1)=0.03(8)\\\Rightarrow f'(x=1)=0.24\ \text{miles per sec}[/tex]
Using derivatives, it is found that the rate of change of elevation at x = 1 is of 0.24.
What is the rate of change of a function f(x)?
- The rate of change of a function f(x) at x = a is given by:
[tex]f^{\prime}(a)[/tex]
In this problem, the function is:
[tex]f(x) = 0.03(10x - x^2)[/tex]
Hence:
[tex]f^{\prime}(x) = 0.3 - 0.06x[/tex]
At x = 1:
[tex]f^{\prime}(1) = 0.3 - 0.06(1) = 0.24[/tex]
The rate of change of elevation at x = 1 is of 0.24.
To learn more about derivatives, you can take a look at https://brainly.com/question/18590720
