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Norma-Jean [14]
3 years ago
11

Computing growth rates (I): Suppose Xt=(1.04)^t and Yt=(1.02)^tCalculate the growth rate of Zt in each of the following cases:(a

) z=xy(b) z=x/y(c) z=y/x(d) z=x^1/2 y^1/2(e) z=(x/y)^2(f) z=x^-1/3y^2/3
Mathematics
1 answer:
Doss [256]3 years ago
5 0

Answer:

z(t) = (0.9808)^t

z(t) = (1.0196)^t

z(t) = (0.9808)^t

z(t) = (1.03)^t

z(t) = (1.0404)^t

z(t) = (1.0001)^t

Step-by-step explanation:

We are given the following in the question:

x(t)=(1.04)^t\\y(t)t=(1.02)^t

We have to find the growth rate z(t) in each of the following cases:

(a) z = xy

z(t) = x(t)y(t)\\z(t) = (1.04)^t.(1.02)^t\\z(t) = (1.04\times 1.02)^t\\z(t) = (1.0608)^t

(b) z=x/y

z(t) =\displaystyle\frac{x(t)}{y(t)}\\\\z(t) = \frac{(1.04)^t}{(1.02)^t} = \bigg(\frac{1.04}{1.02}\bigg)^t\\\\z(t) = (1.0196)^t

(c) z=y/x

z(t) =\displaystyle\frac{y(t)}{x(t)}\\\\z(t) = \frac{(1.02)^t}{(1.04)^t} = \bigg(\frac{1.02}{1.04}\bigg)^t\\\\z(t) = (0.9808)^t

(d) z=x^(1/2) y^(1/2)

z(t) = (x(t))^{\frac{1}{2}}(y(t))^{\frac{1}{2}}\\z(t) = ((1.04)^t)^\frac{1}{2} ((1.02)^t)^\frac{1}{2}\\z(t) = (1.0608)^{\frac{t}{2}}\\z(t) = (1.03)^t

(e) z=(x/y)^2

z(t) =\bigg(\displaystyle\frac{x(t)}{y(t)}\bigg)^2\\\\z(t) =\bigg( \frac{(1.04)^t}{(1.02)^t}\bigg)^2 = \bigg(\frac{1.04}{1.02}\bigg)^{2t}\\\\z(t) = (1.0404)^t

(f) z=x^(-1/3)y^(2/3)

z(t) = (x(t))^{\frac{-1}{3}}(y(t))^{\frac{2}{3}}\\z(t) = ((1.04)^t)^{\frac{-1}{3}}((1.02)^t)^{\frac{2}{3}}\\z(t) = ((1.04)^{\frac{-1}{3}})^t((1.02)^{\frac{2}{3}})^t\\z(t) = (1.04^{\frac{-1}{3}}\times 1.02^{\frac{2}{3}})^t\\z(t) = (1.0001)^t

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The statement that is true about the function is D. it is discontinuous and non-differentiable at x = 3.

<h3>How to determine which statement is true?</h3>

To determine which statement is true, we need to know the conditions for continuity and differentiablity of a function.

<h3>Conditions for continuity and differentiablity of a function.</h3>
  • For a function f(x) to be continuous at a point x = a, then both the left hand limit of f(x) and the right hand limit of f(x) as x → a must be equal. That is \lim_{x \to a^{-} } f(x) =  \lim_{x \to a^{+} } f(x). So,  \lim_{x \to a^{} } f(x) must exist since  \lim_{x \to a^{-} } f(x) =  \lim_{x \to a^{+} } f(x) =  \lim_{x \to a^{} } f(x)
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So, the function is discontinuous and non-differentiable at x = 3.

So, the statement that is true about the function is D. it is discontinuous and non-differentiable at x = 3.

Learn more about continuity of a function here:

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