The number of solutions of equation E ^ x + X-2 = 0

The number of solutions of equation E ^ x + X-2 = 0

y=e^x
y=-x+2
Then the solution of e ^ x + X-2 = 0 is the intersection of the two functions
Y = e ^ x is an increasing function
Y = - x + 2 is a decreasing function
So there's at most one intersection
x=0,e^x-x+2
So there is an intersection at (0,1)
So there is a solution

The number of real roots of the equation x | x | - 3 | x | + 2 = 0 is______ One

① X | x | - 3 | x | + 2 = x2-3x + 2 = 0, (x-1) (X-2) = 0, the solution is x = 1 or 2, satisfying the condition; ② x | 0, X | - 3 | - x | + 2 = - x2 + 3x + 2 = 0, ∫ x2-3x-2 = 0, ∫ Δ = 9-4 × (- 2) = 17, ∫ x = 3 ± 172, ∫ x ＜ 0, ∫ x = 3 − 172, ∫ the number of real roots of the equation is 3, and the answer is 3

This paper discusses the number of real roots of the equation E ^ x + e ^ - x = a, where a is a positive real number

f(x)=e^x+e^(-x)
x∈R
f(x)≥2
So when a > 2, there are two real roots,
When a = 2, there is a real root and x = 0
When a

For the derivative of y = (x ^ 3-x-2 π) / x ^ 2, the answer is 1 + 1 / x ^ 2 + 4 π / x ^ 3

y=x-1/x-2π/x^2
Take the derivative of each term separately
x'=1
(-1/x)'=1/x^2
(-2π/x^2)'=4π/x^3

Y = a ^ x + x ^ A + x ^ x to find the derivative of Y

y=a^x+x^a+x^x
y - a^x - x^a = x^x
ln(y - a^x - x^a) = xlnx
The derivative of X is obtained on both sides
[y' -(a^x)lna - x^(a-1)]/(y - a^x - x^a) = lnx + x*1/x = lnx + 1
y' -(a^x)lna - x^(a-1) = (lnx + 1)(y - a^x - x^a) = (lnx + 1)x^x
y' = (lnx + 1)x^x + (a^x)lna + x^(a-1)

Finding the derivative 2 of y = x ^ 3 + log2x is the subscript

y'=(x^3)'+[log2(x)]'
=3x^2+1/x*1/ln2
=3x^2+1/(xln2)

Find the function y = log2x y = 2ex of the following derivatives

Derivative y = log2x (3-x), x = 1
Help me figure it out
Finding the derivative of y = log2x (3-x) | x = 1

If y = Log &; [x (3-x)] = ln [x (3-x)] / LN2 = [LNX + ln (3-x)] / ln2dy / DX = 1 / (xln2) - 1 / (3-x) LN2 if x = 1dy / DX | (x = 1) = 1 / LN2 - 1 / 2ln2 = 1 / 2ln2 (or 1 / ln4) if y = log (2x) [3-x] = ln (3-x)

What is the derivative of y = e ^ (x ^ 2)
Is it 2xe ^ (x ^ 2) or 4xe ^ (x ^ 2). If it's the former, why don't you continue to seek after 2x

Derivative of y = x ^ 2 + e ^ x + SiNx
It's better to have steps

y=x^2+e^x+sinx
The derivative 2x of x ^ 2
Derivative e ^ X of e ^ x
The derivative cosx of SiNx
So the derivative of y = x ^ 2 + e ^ x + SiNx is 2x + e ^ x + cosx