Question

F(x)=(4x-3)^2 в х0=2

Answer 1

Решение на прикреплённой фотографии

Answer 2

Відповідь:    у = 40х - 55 - рівняння дотичної .

Пояснення:

     f(x) = ( 4x -3 )² ;   х₀=2 ;   y = f( x₀ ) + f '( x₀ )( x - x₀ ) - р - ня дотичної

  f( x₀ ) = ( 4*2 - 3 )² = 5² = 25 ;

  f '( x ) = [ ( 4x -3 )² ]' = 2( 4x - 3 )*( 4x - 3 )' = 2( 4x - 3 )* 4 = 8( 4x - 3 ) ;

  f '( x₀ ) = 8( 4*2 - 3 ) = 8* 5 = 40 .  Підставимо значення :

   у = 25 + 40( х - 2 ) = 25 + 40х - 80 = 40х - 55 ;   у = 40х - 55 .

Answer 3

To find the derivative of the function $F(x) = (4x - 3)^2$ at the point $x_0 = 2$, you can use the power rule and chain rule. Here's how you can do it step by step:

1. Find the derivative of $F(x)$ with respect to $x$ using the power rule:

   $F'(x) = 2(4x - 3)^{2-1} \cdot 4 = 8(4x - 3)$

2. Now, evaluate $F'(x)$ at the point $x_0 = 2$:

   $F'(2) = 8(4 \cdot 2 - 3) = 8(8 - 3) = 8 \cdot 5 = 40$

So, the derivative of $F(x) = (4x - 3)^2$ at $x = 2$ is $F'(2) = 40$.

Answer 4

To find the derivative of the function $f(x) = (4x - 3)^2$ at the point $x_0 = 2$, you can use the power rule and the chain rule for differentiation. Here's how you can do it step by step:

1. Start with the function $f(x) = (4x - 3)^2$.

2. Apply the chain rule, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. In this case, the inner function is $4x - 3$, and the outer function is squaring.

3. Find the derivative of the inner function: $f'(x) = 2(4x - 3)^{2-1} \cdot \frac{d}{dx}(4x - 3)$

4. Find the derivative of $4x - 3$: $\frac{d}{dx}(4x - 3) = 4$

5. Plug this result back into the chain rule expression: $f'(x) = 2(4x - 3) \cdot 4$

6. Now, you have the derivative of $f(x)$: $f'(x) = 8(4x - 3)$

7. To find the derivative at the point $x_0 = 2$, substitute $x = 2$ into the derivative expression: $f'(2) = 8(4(2) - 3)$ $f'(2) = 8(8 - 3)$ $f'(2) = 8(5)$ $f'(2) = 40$

So, the derivative of the function $f(x) = (4x - 3)^2$ at $x = 2$ is $f'(2) = 40$.