Question

Найдите f'(6), если f(x)=|x²-14x+45|

Answer 1

sign(x) = -1, x<0

sign(x) = 1, x > 0

sign(x) = 0, x = 0

$\frac{d|u|}{x} =sign(u)\frac{du}{dx} \\ u=x^2-14x+45\\ u'(x)=2x-14\\ f'(6)=sign(f(6))*u'(6)\\ f(6)=36-84+45=-3\\ sign(f(6))=-1\\ u'(6)=12-14=-2\\ f'(6)=-1*(-2)=2$

Answer 2

Ответ на картинке внизу страницы

Answer 3

To find the derivative of the function f(x) = |x² - 14x + 45| and then evaluate it at x = 6, we need to consider the two cases when x² - 14x + 45 is positive and negative.

Step 1: Find the derivative of f(x) for x² - 14x + 45 > 0 (i.e., when x is outside the roots of the quadratic).

In this case, |x² - 14x + 45| = x² - 14x + 45, so we need to find the derivative of x² - 14x + 45.

f(x) = x² - 14x + 45 f'(x) = d/dx (x²) - d/dx (14x) + d/dx (45) f'(x) = 2x - 14

Step 2: Find the derivative of f(x) for x² - 14x + 45 < 0 (i.e., when x is between the roots of the quadratic).

In this case, |x² - 14x + 45| = -(x² - 14x + 45), so we need to find the derivative of -(x² - 14x + 45).

f(x) = -(x² - 14x + 45) f'(x) = -d/dx (x²) + d/dx (14x) - d/dx (45) f'(x) = -2x + 14

Step 3: Determine the interval for which x² - 14x + 45 > 0 or x² - 14x + 45 < 0.

The roots of x² - 14x + 45 = 0 can be found by setting the quadratic expression equal to zero and solving for x:

x² - 14x + 45 = 0 (x - 5)(x - 9) = 0

The roots are x = 5 and x = 9.

Now, we can determine the intervals:

1. When x < 5: x² - 14x + 45 > 0
2. When 5 < x < 9: x² - 14x + 45 < 0
3. When x > 9: x² - 14x + 45 > 0

Step 4: Evaluate f'(x) at x = 6.

Since 5 < 6 < 9, we use the second derivative we found for the interval 5 < x < 9:

f'(6) = -2(6) + 14 f'(6) = -12 + 14 f'(6) = 2

So, f'(6) = 2 when x = 6.