Using the differential mean value theorem to prove that the equation X5 + X-1 = 0 has only one positive root?

Using the differential mean value theorem to prove that the equation X5 + X-1 = 0 has only one positive root?

Using the differential mean value theorem to prove that there are only three different real roots of an equation
Using the differential mean value theorem to prove the equation: 2 ^ x-x ^ 2-1 = 0 has and only has three different real roots on the whole number axis

It is proved that the equation: 2 ^ x-x ^ 2-1 = 0 has and only has three different real roots on the whole number axis. It is proved that: y = f (x) = 2 ^ x-x ^ 2-1. Obviously, f (0) = f (1) = 0, F & acute; (x) = (LN2) * (2 ^ x) - 2x, F & acute; (0) = LN2, F & acute; (1) = - 2 (1-ln2), F "(x) = (LN2) & sup2; * (2 ^ x) - 2, so that f" (x) = 0, we can turn

If 3A ^ 2-5b

First, 3A ^ 20
Then we get the derivative, and we get 5x ^ 4 + 6AX ^ 2 + 3B. Obviously, the derivative is always greater than or equal to zero
So the equation has at most one real root on a real number
It is also noted that when x tends to negative infinity, the value of the function tends to negative infinity;
When x tends to positive infinity, the function value also tends to positive infinity
therefore
There is a unique real root