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resubmission of q31

Text page 45 Chapter 1 Section 4 Problem #31.

I have resubmitted your response with a clarification and another question inserted.

Question 31 chapter 1 section4

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Evaluate the difference quotient and simplify the result.

31 g(x)= sq. rt. of x+1

g(x+ delta x) - g(x)/ delta x

You aren't consistent in using parentheses where they are needed. So I can't tell whether the function is sqrt(x+1) or sqrt(x) + 1.

If it is the former then the 1 has to stay within the parentheses. If the latter your expression is OK.

???? g(x) = sq. root of (x+1) is the correct expression so could you please rework it for me with that?????

Assuming that your expression means sqr(x) + 1, then

sq. rt. of (x+delta x) +1 - sq. rt. of x+1

should be written

sq. rt. of (x+delta x) +1 - (sq. rt. of x+1)

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Using the solution on Calc Chat

I understand that I am subbing sq. rt. of x+1 into the numerator to get:

What I do not understand is

1. Why did I not sub. sq. root of x+1 into the denominator also?

In the first step the denominator is delta x. It seems that it should be delta sq. root of x.

The difference quotient represents an average rate of change of f(x) with respect to x, which by definition is (change in f(x)) / (change in x).

The numerator of the difference quotient is the change in the value of the function, the denominator is the change in the value of x.

The numerator therefore represents the change that results in the value of the function when x changes by `dx, and the denominator is just `dx, which is the change in x.

So the denominator will always be `dx.

???? Is this also why when you plug in g(x) = sq. root of (x+1) into the numerator, the Calc Chat solution shows

g(sq root of (x+ change in x +1)) - (sq. root of (x+1))

instead of

g(sq root of (x+1)) + change in (sq. root of (x+1)) - g( sq. root of (x+1)) in the denominator?

So the change in x, wherever it occurs, does not change - nothing is substituted for it? ????

2.Why was both the numerator and denominator multiplied by sq. rt. of x+ delta x +1 + sq. rt. of x +1?

I thought an expression was simplified if there were no radicals in the denominator. This takes the radicals out of the numerator and puts them in the denominator.

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I need more notes on this. I am not following the one example in the book at all, and there is no explanation of the steps on Calc Chat.

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The goal isn't to simplify the expression, it is to find the limiting value of the expression as `dx approaches zero. The multiplication of the denominator accomplishes this.

If the parentheses are as I'm assuming, the numerator

sq. rt. of (x+delta x) +1 - (sq. rt. of x+1)

becomes

sq. rt. of (x+`dx) +1 - sq. rt. of x-1

which is equal to

sqrt(x+`dx) - sqrt(x).

Multiplying numerator and denominator by sqrt(x+`dx) + sqrt(x) gives you

(x + `dx - x) / (`dx * (sqrt(x+`dx) + sqrt(x) ) )

which is just equal to

`dx / (`dx * (sqrt(x+`dx) + sqrt(x) ) ),

which in turn simplifies to

1 / (sqrt(x+`dx) + sqrt(x) ).

As `dx approaches zero, this approaches

1 / (sqrt(x) + sqrt(x) ) = 1 / (2 sqrt(x) ).

This is a challenging problem and you're doing very well by asking good questions about it.

Do get in the habit of being very careful with your use of parentheses to indicate grouping.

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g(x+`dx) = sqrt(x+`dx + 1) so

(g(x+`dx) - g(x) ) / `dx

= ( sqrt(x+`dx+1) - sqrt(x+1) ) / `dx

= ( sqrt(x+`dx+1) - sqrt(x+1) ) / `dx * ( sqrt(x+`dx+1) + sqrt(x+1) ) / ( sqrt(x+`dx+1) + sqrt(x+1) )

= ( (x + `dx + 1) - (x+1) ) / (`dx ((sqrt(x+`dx+1) + sqrt(x+1) ) )

= `dx / / (`dx ((sqrt(x+`dx+1) + sqrt(x+1) ) )

= 1 / ( ((sqrt(x+`dx+1) + sqrt(x+1) ) ).

As `dx -> - this approaches

= 1 / ( ((sqrt(x+1) + sqrt(x+1) ) ),

which is the same as

1 / (2 sqrt(x+1) )
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Question about major quiz

I noticed when I went to take the Major Quiz on Saturday that the directions said a graphing calculator was permissible. I thought in the initial information that it stated that the only permissible calculator for tests was the Casio fx-260, which is not a graphing calculator. Did I misunderstand, or is the graphing calculator permissible on quizzes but not tests?

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@& The instructions on the test were incorrect. However a graphing calculator would have made no difference on the Major Quiz. *@

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The instructions on the test were incorrect.

However a graphing calculator would have made no difference on the Major Quiz.
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