Question

Suppose that s(x)=e−μˣ ,x⩾0, Find eₓ and e˙ₓ​ .

Answer 1

To find eₓ and e˙ₓ for the function s(x) =
`e^(-μx),`eₓ is simply the function itself, which is
`e^(-μx).`

`eₓ = s(x) = e^(-μx)`

`e˙ₓ = s'(x) = -μe^(-μx)`

Step-by-step explanation:

To find eₓ and e˙ₓ for the function s(x) =
`e^(-μx)`, eₓ is simply the function itself, which is
`e^(-μx)`.

This represents the value of the function s(x) at a given x. For e˙ₓ, it requires taking the derivative of s(x) with respect to x.

The derivative of
`e^(-μx)`with respect to x is found using the chain rule.

The derivative of
`e^(-μx)` is
`-μe^(-μx)`, where -μ is the constant multiplied by the original function,
`e^(-μx)`, and the negative sign indicates the function's decreasing behavior as x increases.

So,

for
`s(x) = e^(-μx):`

`eₓ = e^(-μx)`

`e˙ₓ = s'(x) = d/dx(e^(-μx)) = -μe^(-μx)`

These expressions showcase how eₓ represents the original function itself, while e˙ₓ represents the rate of change or the derivative of the function with respect to x.

The negative sign in e˙ₓ indicates that the function is decreasing as x increases due to the negative exponential term.

This differentiation helps in understanding the behavior of the function over the given domain, particularly in modeling scenarios where exponential decay or decreasing trends are involved.

Answer 2

To find eₓ and e˙ₓ for the function s(x) = e^(-μx), eₓ is simply the function itself, which is e^(-μx).

eₓ = s(x) = e^(-μx)

e˙ₓ = s'(x) = -μe^(-μx)

Step-by-step explanation:

To find eₓ and e˙ₓ for the function s(x) = e^(-μx), eₓ is simply the function itself, which is e^(-μx). This represents the value of the function s(x) at a given x. For e˙ₓ, it requires taking the derivative of s(x) with respect to x. The derivative of e^(-μx) with respect to x is found using the chain rule. The derivative of e^(-μx) is -μe^(-μx), where -μ is the constant multiplied by the original function, e^(-μx), and the negative sign indicates the function's decreasing behavior as x increases.

So,

for s(x) = e^(-μx):

eₓ = e^(-μx)

e˙ₓ = s'(x) = d/dx(e^(-μx)) = -μe^(-μx)

These expressions showcase how eₓ represents the original function itself, while e˙ₓ represents the rate of change or the derivative of the function with respect to x. The negative sign in e˙ₓ indicates that the function is decreasing as x increases due to the negative exponential term.

This differentiation helps in understanding the behavior of the function over the given domain, particularly in modeling scenarios where exponential decay or decreasing trends are involved.