Question

Let fx=√x. If the rate of change of f at x=c is twice its rate of change at x=1, then c=__? A. 1/4 B. 1 C. 4 D. 1√2 E. 1/(2√ 2)

Answer 1

The value of c that makes the rate of change of f(x) at x=c twice that at x=1 for the function f(x)=√x is c=1/4.

Step-by-step explanation:

The student is asking for the value of c that makes the rate of change of the function f(x) = √x at x = c twice the rate of change at x = 1. To find the rate of change, we need to calculate the derivative of f(x), which is f'(x) = (1/2)x^{-1/2}. Evaluating this derivative at x = 1, we get f'(1) = 1/2. To find the value of c where the rate of change is double this amount, we set f'(c) = 2 × (1/2) and solve for c. After simplification, this gives us f'(c) = 1, and upon setting this equal to (1/2)c^{-1/2} and then solving for c, we find that c = 1/4.

Answer 2

So the answer is A. 1/4.

Of course, I've been enhancing my skill in dealing with derivative problems. Let's find the value of c for which the rate of change of the function f(x) = √x at x = c is twice its rate of change at x = 1.

We are given that:

f(x) = √x

f'(x) = 1/(2√x)

f'(c) = 2 * f'(1)

We can solve this system of equations to find the value of c.

Steps to solve:

1. Differentiate f(x):

f'(x) = 1/(2√x)

2. Substitute the expressions for f'(x) in both equations:

1/(2√c) = 2 * 1/(2√1)

3. Simplify:

1/(2√c) = 2

4. Solve for c:

√c = 1/2

c = (1/2)^2

c = 1/4

Therefore, the value of c for which the rate of change of f(x) at x = c is twice its rate of change at x = 1 is 1/4.