The number of real roots of equation x ^ 3-6x ^ 2 + 9x-10 = 0 is

Let f (x) = x ^ 3-6x ^ 2 + 9x-10
‡ derivation f (x) '= 3x ^ 2-12x + 9 = 3 (x-3) (x-1)
When x = 3 and x = 1, f (x) has extreme value f (1) = - 6

The number of real roots of equation x ^ 3-6x ^ 2 + 9x-10 = 0 is () A.3 B.2 C.1 D.O

Choose C
Let f (x) = x ^ 3-6x ^ 2 + 9x-10
When x < 1, f (x) monotonically increases, f (1) = - 6 < 0
When 1 < x < 3, f (x) decreases monotonically, f (1) = - 6, f (3) = - 10
When x > 3, f (x) monotonically increases, f (3) = - 10 < 0
F (+ ∞) > 0, so the function has a root in (3, + ∞)

A problem on the derivative of Higher Mathematics Let f (x) be differentiable, f (0) = 1, f '(- LNX) = x, f (1) =?

f'(-lnx)=x
Get f '(x) = e ^ (- x)
Then the indefinite integral f (x) = - e ^ (- x) + C
Because f (0) = - 1 + C = 1
C=2
f（x）=-e^(-x)+2
f（1）=-1/e+2

For a quadratic equation of one variable, the product of two is equal to And Coefficient___

For a quadratic equation of one variable, the product of two is equal to the ratio of (constant term) to (quadratic term) coefficients

Given that the quadratic term coefficient of a univariate quadratic equation is 2, the sum of its two is 3, and the product of the two is - 2, find the univariate quadratic equation

Let this univariate quadratic equation be 2x2 + PX + q = 0,
∵ the sum of the two is 3, and the product of the two is - 2,
∴-p
2=3，q
2=-2，
p=-6，q=-4，
The equation is: 2x2-6x-4 = 0

What is the sum of two quadratic equations of one variable and what is the product of two equations?

The coefficient of the quadratic term is B, the constant term is C, and the sum of the two is equal to - A / b \ x0d, and the product of the two is equal to a / C. I hope to adopt the supplement:

The number of real roots of equation x ^ 3-6x ^ 2 + 9x-10 = 0 is