Question

Решите уравнение f'(x)=0,если: a)f(x)=4x^2-8x+5 б)f(x)=3x^3-8x^2+7x-2

Answer 1

A) f'(X) =8x-8
8x-8=0
X=1
B)f'(x)=9x^2-16x+7
9x^2-16x+7=0
D=256-252=4
X1=1
X2=7/9

Answer 2

To find the critical points of the given functions, we need to find the values of 'x' for which the derivative of each function is equal to zero (f'(x) = 0).

a) For the function f(x) = 4x^2 - 8x + 5:

Step 1: Find the derivative of f(x): f'(x) = d/dx(4x^2 - 8x + 5) = 8x - 8

Step 2: Set f'(x) equal to zero and solve for 'x': 8x - 8 = 0 8x = 8 x = 1

b) For the function f(x) = 3x^3 - 8x^2 + 7x - 2:

Step 1: Find the derivative of f(x): f'(x) = d/dx(3x^3 - 8x^2 + 7x - 2) = 9x^2 - 16x + 7

Step 2: Set f'(x) equal to zero and solve for 'x': 9x^2 - 16x + 7 = 0

Now, we can use the quadratic formula to solve for 'x':

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 9, b = -16, and c = 7.

x = (16 ± √((-16)^2 - 4 * 9 * 7)) / (2 * 9) x = (16 ± √(256 - 252)) / 18 x = (16 ± √4) / 18 x = (16 ± 2) / 18

Two possible solutions for 'x' are:

1. x = (16 + 2) / 18 = 18 / 18 = 1
2. x = (16 - 2) / 18 = 14 / 18 = 7 / 9 ≈ 0.778

So, for function b), the critical points are x = 1 and x ≈ 0.778.

To summarize: a) The critical point for f(x) = 4x^2 - 8x + 5 is x = 1. b) The critical points for f(x) = 3x^3 - 8x^2 + 7x - 2 are x = 1 and x ≈ 0.778.