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2 years ago

What is the average rate of change of the function on the interval from x = 0 to x = 5?

Mathematics

2 answers:

bixtya [17]2 years ago

8 0

The average rate is 3^x/2

damaskus [11]2 years ago

6 0

Answer:

Average rate of change = 24.2

Step-by-step explanation:

$f(x)=\frac{1}{2}3^x$

Average rate of change formula is on interval a to b is

$\frac{f(b)-f(a)}{b-a}$

Given interval is x=0 to x=5

a=0 and b=5

Average rate = $\frac{f(5)-f(0)}{5-0}$

$f(0)=\frac{1}{2}3^0=\fracc{1}{2}=0.5$

$f(5)=\frac{1}{2}3^5=\frac{3^5}(2}=121.5$

Average rate = $\frac{f(5)-f(0)}{5-0}$

Average rate = $\frac{121.5-0.5}{5}=24.2$

Average rate of change = 24.2

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x≤-2

Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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17. El precio de los terrenos en Lima es proporcional al área e inversamente proporcional a su distancia con respecto al centro

grin007 [14]

Answer:

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Step-by-step explanation:

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3 years ago

Find the indicated limit, if it exists.

kondor19780726 [428]

Answer:

d) The limit does not exist

General Formulas and Concepts:

Calculus

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$
Left-Side Limit: $\displaystyle \lim_{x \to c^-} f(x)$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

Step-by-step explanation:

*Note:

In order for a limit to exist, the right-side and left-side limits must equal each other.

Step 1: Define

Identify

$\displaystyle f(x) = \left\{\begin{array}{ccc}5 - x,\ x < 5\\8,\ x = 5\\x + 3,\ x > 5\end{array}$

Step 2: Find Right-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^+} 5 - x$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} 5 - x = 5 - 5 = 0$

Step 3: Find Left-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^-} x + 3$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} x + 3 = 5 + 3 = 8$

∴ Since $\displaystyle \lim_{x \to 5^+} f(x) \neq \lim_{x \to 5^-} f(x)$ , then $\displaystyle \lim_{x \to 5} f(x) = DNE$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

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2 years ago