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Problem 16- Solution -pg. 3/3

Posted in Uncategorized by Lia on 06/16/2011

Problem 16 – Solution: 12, 3

4. Filling the empty spaces in VT.

4.1 The signs for f{prime}(x)

From point 2.3 of this presentation we can see that:

f{prime}(x) = {{x - 2}/{sqrt{{x^2} - 4x + 3}}}

The denominator is positive on the defined interval (it is equal with f(x)) therefore the sign of f{prime}(x) is dependent on the sign of its numerator.

f{prime}(x) = o when x = 2.

x = 2 would have been a minimum if it was in a defined domain.

When {x < 2} {right}{f{prime}(x) < o}

When {x > 2} {right}{f{prime}(x) > o}” title=”{x > 2} {right}{f{prime}(x) > o}”/><img src=

4.2 Input {}{nearrow}{} and {}{searrow}{} on f(x) row :

  • {}{nearrow}{} for f(x) when f{prime}(x) is “+

  • {}{searrow}{} for f(x) when f{prime}(x) is “-

5. Graphing, on Cartesian Coordinates System the function.

5.1 Drawing the oblique asymptotes  calculated at point 3.1 in this presentation.

5.2 Drawing all set points (x, f(x)):

(0, sqrt{3}) and (1,0), (3,0) and with approximation the points where x or f(x) is {}{-{infty}}{} or {}{+{infty}}{}.

5.3 Graphing the function f(x)(Fig. 5).

 

 

Problem 16 – Solution: 12, 3