For the function f(x) = ln(2x+3) - 2x at x = 1:
The derivative f'(x) is: f'(x) = 2/(2x+3) - 2
Evaluating at x = 1, we get:
f'(1) = 2/5 - 2 = -6/5
and f(1) = ln(5) - 2
Elasticity E = 1 * (-6/5) / (ln(5) - 2)
For the function f(x) = sqrt(x^2+4) at x = -2:
The derivative f'(x) is: f'(x) = x/sqrt(x^2+4)
Evaluating at x = -2, we get:
f'(-2) = -2/sqrt(8) = -2/(2*sqrt(2)) = -1/sqrt(2)
and f(-2) = sqrt(8) = 2*sqrt(2)
Elasticity E = -2 * (-1/sqrt(2)) / (2*sqrt(2))
For the function f(x) = ln(x+3) at x = 0:
The derivative f'(x) is: f'(x) = 1/(x+3)
Evaluating at x = 0, we get:
f'(0) = 1/3
and f(0) = ln(3)
Elasticity E = 0 * 1/3 / ln(3)
For the function f(x) = 4x at x = 2:
The derivative f'(x) is: f'(x) = 4
Evaluating at x = 2, we get:
f'(2) = 4
and f(2) = 8
Elasticity E = 2 * 4 / 8 = 1
Using the calculated elasticity values:
If E = 1, it's Unit elastic.
If E < 1 and E > 0, it's Inelastic.
If E = 0, it's Perfectly inelastic.
If E > 1, it's Elastic.
For f(x) = ln(2x+3) - 2x at x = 1:
Elasticity E = 1 * (-6/5) / (ln(5) - 2)
Without calculating the exact value, it's clear that E will be negative, which doesn't fit the standard interpretation of elasticity. We'll skip categorization for this function due to its negative value.
For f(x) = sqrt(x^2+4) at x = -2:
Elasticity E = -2 * (-1/sqrt(2)) / (2*sqrt(2))
This simplifies to E = 0.5 which is less than 1.
This is Inelastic.
For f(x) = ln(x+3) at x = 0:
Elasticity E = 0 * 1/3 / ln(3)
This simplifies to E = 0.
This is Perfectly inelastic.
For f(x) = 4x at x = 2:
Elasticity E = 2 * 4 / 8
This simplifies to E = 1.
This is Unit elastic.
So, in summary:
f(x) = ln(2x+3) - 2x at x = 1: Cannot be categorized using the given options due to a negative elasticity value.
f(x) = sqrt(x^2+4) at x = -2: Inelastic.
f(x) = ln(x+3) at x = 0: Perfectly inelastic.
f(x) = 4x at x = 2: Unit elastic.
