What is minimum value of e^2x+ e^-2x
Thx
recall that this is just 2 cosh(x).
d/dx (cosh x) = sinh(x)
sinh(x) = 0 at x=0
so, the minimum value is 1+1 = 2
(or 2cosh(0) = 2)
Or, if you insist on the hard way,
y' = 2e^2x - 2e^-2x
= 2e^-2x (e^4x - 1)
y'=0 at e^4x=1, or x=0.
To find the minimum value of the expression e^2x + e^-2x, we can use the concept of calculus and the properties of exponential functions.
Let's consider the function y = e^2x + e^-2x. To find its minimum value, we need to find the critical points by taking the derivative and setting it equal to zero.
Step 1: Find the derivative of y with respect to x.
dy/dx = (d/dx)(e^2x) + (d/dx)(e^-2x)
Using the chain rule, the derivative of e^2x with respect to x is 2e^2x, and the derivative of e^-2x with respect to x is -2e^-2x.
So, dy/dx = 2e^2x - 2e^-2x.
Step 2: Set dy/dx = 0 and solve for x.
2e^2x - 2e^-2x = 0
Divide both sides of the equation by 2:
e^2x - e^-2x = 0
Multiply both sides of the equation by e^2x:
(e^2x)^2 - 1 = 0
Using the identity (a^2 - b^2) = (a - b)(a + b), we can factor the left side of the equation:
(e^2x - 1)(e^2x + 1) = 0
Now, set each factor equal to zero:
e^2x - 1 = 0 or e^2x + 1 = 0
Solving the first equation:
e^2x = 1
Taking the natural logarithm of both sides:
ln(e^2x) = ln(1)
2x ln(e) = 0
2x = 0
x = 0
Solving the second equation:
e^2x = -1
Since e^2x is always positive, there is no solution for this equation.
Therefore, the critical point is x = 0.
Step 3: Determine whether the critical point is a minimum or maximum by analyzing the second derivative.
To do this, we need to find the second derivative of y with respect to x.
d^2y/dx^2 = (d/dx)(2e^2x - 2e^-2x)
= 4e^2x + 4e^-2x
Substitute x = 0 into the second derivative:
d^2y/dx^2 = 4e^0 + 4e^0
= 4 + 4
= 8
Since the second derivative, 8, is positive, we can conclude that x = 0 is a minimum point.
Step 4: Find the minimum value of y by substituting x = 0 back into the original expression.
y = e^2(0) + e^-2(0)
y = 1 + 1
y = 2
Therefore, the minimum value of e^2x + e^-2x is 2.