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Suppose the graphs of $ f(x) = x^2 $ and $ g(x) = 2^x $ are drawn on a coordinate grid where the unit of measurement is 1 inch. Show that, at a distance 2 ft to the right of the origin, the height of the graph of $ f $ is 48 ft but the height of the graph of $ g $ is about 265 mi.

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$2 \mathrm{ft}=24$ in, $f(24)=24^{2}$ in $=576$ in $=48 \mathrm{ft}$ $g(24)=2^{24}$ in $=2^{24} /(12 \cdot 5280) \mathrm{mi} \approx 265 \mathrm{mi}$

03:30

Jeffrey Payo

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 4

Exponential Functions

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

Johns Hopkins University

Missouri State University

Campbell University

Boston College

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all right, So if we're graphing these functions on agreed where one unit is one inch and we've gone over two feet to the right of the origin. That means we've gone over 24 inches to the right of the origin. We would be at X equals 24. So what we want to find is the height of the point F of 24 the height of the point G of 24 compare those heights. So let's find F of 24 F of 24 would be 24 squared and 24 squared is 576. That would be 576 inches if we're going by. One unit is one inch now 576 inches is equivalent to 48 feet. All right, now, let's find out what G of 24 is, so that would be to raise to the 24 power. And for that we get 16 million 777,000 216 inches. Okay, let's figure out how many feet that would be. We divide that by 12 and we get oh, about 1,398,000 feet. And if we want to know how many miles that is, we divide that by 5280 and we get about 265 miles approximately so. The point of that is to show us how much faster the exponential function grows compared to the polynomial function, the Y values get much, much bigger, and it doesn't take very long for that to happen.

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