Question

При каких значениях х выполняется равенство f' (x) = f (x) , если f (x) =(2x - 1)^2 ?

Answer 1

Ответ:

f(x)=x^2+3/vX

f'(x)=2x-3/2*x^(-2/3)

2x-3/2*(x^(-2/3)=0

ОДЗ. x>0

2x=3/2*(x^(2/3)

x^(5/3)=0,75

x=5V0,421875

Answer 2

To find the values of x for which the equation f'(x) = f(x) holds true, let's first differentiate the function f(x) = (2x - 1)^2.

Using the chain rule, we can differentiate f(x) as follows:

f'(x) = 2(2x - 1)(2) = 4(2x - 1).

Now, let's set up the equation f'(x) = f(x):

4(2x - 1) = (2x - 1)^2.

Expanding the right side of the equation:

4(2x - 1) = 4x^2 - 4x + 1.

Simplifying:

8x - 4 = 4x^2 - 4x + 1.

Rearranging the equation to form a quadratic equation:

4x^2 - 12x + 5 = 0.

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a),

where a = 4, b = -12, and c = 5.

Plugging in the values:

x = (12 ± √((-12)^2 - 4(4)(5))) / (2(4)).

x = (12 ± √(144 - 80)) / 8.

x = (12 ± √64) / 8.

x = (12 ± 8) / 8.

Now, we have two possible solutions for x:

1. x = (12 + 8) / 8 = 20 / 8 = 2.5.
2. x = (12 - 8) / 8 = 4 / 8 = 0.5.

Therefore, the values of x for which the equation f'(x) = f(x) holds true in the given function f(x) = (2x - 1)^2 are x = 2.5 and x = 0.5.