Question

Let f(x)=2x^(3)-5x^(2)+8x+1 be the function given by on the closed interval [1,3]. What is the instantaneous rate of change of f'(x) x=1?

Answer 1

The instantaneous rate of change of the function f(x) at x=1 is found by calculating its derivative f'(x) and substituting x=1 into it, which results in f'(1) = 4.

Step-by-step explanation:

To find the instantaneous rate of change of the function f(x)=2x3-5x2+8x+1 at x=1 within the closed interval [1,3], we must calculate the derivative of f(x), which is f'(x). The derivative represents the instantaneous rate of change of the function at any point x.

The derivative f'(x) of the function f(x) is calculated as:

Power Rule: To differentiate xn, where n is a constant, multiply by the power and subtract one from the exponent.

Applying the power rule to each term in the function, we get:

• 
  
• f'(x) = d/dx (2x3) - d/dx (5x2) + d/dx (8x) + d/dx (1)
• 
  
• f'(x) = 6x2 - 10x + 8

Now, we substitute x=1 into f'(x) to get the instantaneous rate of change at that point:

• 
  
• f'(1) = 6(1)2 - 10(1) + 8
• 
  
• f'(1) = 6 - 10 + 8
• 
  
• f'(1) = 4

Therefore, the instantaneous rate of change of f(x) at x=1 is 4.

Answer 2

The instantaneous rate of change of f'(x) at x=1 is 4.

Step-by-step explanation:

A function's derivative at a particular location indicates the function's instantaneous rate of change at that location. Finding the instantaneous rate of change of f'(x) at x=1 is the task at hand in this instance. Using the power rule on each term in the polynomial function f(x), we first determine the derivative of f(x) in order to accomplish this.

To find the instantaneous rate of change of f'(x) at x=1, we need to first find the derivative of f(x).

The derivative of
`f(x)=2x^(3)-5x^(2)+8x+1` is
`f'(x)=6x^(2)-10x+8.`

We can then substitute x=1 into f'(x) to find the value of the instantaneous rate of change:

`= f'(1)=6(1)^(2)-10(1)+8`

=4.