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| PAGE: 116 | SET: Exercises | PROBLEM: 85 |
(a)
The function is 
and solution point is 
.
Graph.
Graph the function 
.
Plot the point 
.
Observe the graph.
Consider a point on the curve such that the graph appears to be linear.
One such point is 
.
Find the secant line equation.
The two points are and 
.
Slope of the secant line is 
.
Slope of the secant line is 
.
Point-Slope form of line equation: 
.
Substitute and 
in point-slope form.
Secant line equation is 
.
(b)
Equation of the tangent line is 
.
Consider 
.
Apply derivative on each side with respect to 
.
Substitute 
in 
.
The point is which means that 
.
Tangent line equation is 
.
Substitute 
and 
in the tangent line equation.
Tangent line equation is 
.
Secant line equation is .
The secant line and tangent line appears to be same when the two points come closer.
Hence, the slope of the secant line approaches to tangent line at , as points come closer to .
(c)
Graph the function 
.
Graph the tangent line .
The tangent line 
is the most accurate tangency point.
If the point of the tangency is moved, the approximation will become less accurate.
(d)
Complete the table.
The function is .
The tangent line equation is .
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Observe the table.
We can conclude that as the point moves away, the accuracy of the approximation becomes less.
(a)
The graph is
Approximating point is .
Secant line equation is .
(b)
The slope of the secant line approaches to tangent line at , as points come closer to .
(c)
Graph of the function and tangent line is
If the point of the tangency is moved, the approximation will become less accurate.
(d)
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As the point moves away, the accuracy of the approximation becomes less.
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