It is proved that the third power of equation x - 3x-1 = 0 has at least one root in the interval [- 1,0]

It is proved that the third power of equation x - 3x-1 = 0 has at least one root in the interval [- 1,0]

f(x)=x^3-3x-1,f(-1)=-1-3*(-1)-1=1>0,f(0)=-1

It is proved that there is at least one real root less than 1 between 1 and 2 for the 5th power - 3x + 1 = 0 of equation X

Let f (x) = x ^ 5-3x + 1, then f (x) is continuous on [1,2]
∵ f (1) = - 1 < 0, f (2) = 27 > 0, that is, f (1) and f (2) have different signs
There is at least one real root between [1,2]

It is proved that the 5th power of equation x - the 3rd power of 3x + 1 = 0 has at least one even root in de armour (0,1)

f(x)=x^5-3x ³+ one
f(0)=1>0；
f(1)=-1

It is proved that the equation x * (x power of E) = 1 has at least one positive root less than 1?

prove:
Let f (x) = Xe ^ X-1
Then f (0) = - 10
According to the intermediate value theorem of continuous function, it is known that there must be a ∈ (0,1) so that f (a) = 0

It is proved that the equation x3-3x + C = 0 has at most one real root on [0,1]

It is proved that if f (x) = x3-3x + C, then f '(x) = 3x2-3 = 3 (x2-1)
When x ∈ (0,1), f '(x) ＜ 0 is constant
‡ f (x) decreases monotonically on (0, 1)
The image of F (x) has at most one intersection with the X axis
Therefore, the equation x3-3x + C = 0 has at most one real root on [0,1]

How to prove that the 5th power of equation x minus 3x is equal to 1, and at least one root is between 1 and 2

Find the reciprocal and judge that the monotonicity of the formula of subtracting 3x to the 5th power of X is monotonically increasing
The formula is less than 0 when x = 1 and greater than 0 when x = 2
So there must be a number between 1 and 2 to make it equal to 0
That is, the root is 0
If I didn't learn derivative, when I didn't say it