Basic Understanding

[nosit]

Hello,

In this problem attached, I see that "x" is replaced by "x+h" (it is highlighted in yellow in the picture).

Is there someone able to help me to understand why this substitution happens?

Thank you in advance.

 

[Steven G]

Did you notice the x+h which is highlighted here: f(x+h) - f(x)

I take it that you do not understand what f(x) = 2x - 5 means. Since x is between the parenthesis in f(x) and f(x) = 2x-5, then the job for f is to double whatever is between the parenthesis and subtract 5.

For example, f(12) = 2(12)-5. I replaced x with 12. f(-2/3) = 2(-2/3) - 5. So why not f(x+h) = 2(x+h)-5? After all, I doubled what is between the parenthesis and then subtracted 5.

 

[Steven G]

For the record, 2h/h is NOT 2.
Who ever told you that is incorrect.

 

[nosit]

@Jomo, cristal clear your explanation. thank you.

Regarding your second comment, why it is not 2?

 

[Steven G]

I am glad that my response was clear.

If h = 0, then 2*0/0 is NOT 2. It is undefined!


To be clear, [MATH]\dfrac{x}{x}\neq 1, [/MATH] [MATH]\dfrac{x}{x} = 1[/MATH] if [MATH]x\neq0[/MATH] while[MATH]\dfrac{x}{x}[/MATH]is undefined if x=0

 

[nosit]

Well... I thought (maybe wrongly) that h represents a limit, right? So in theory h will get VERY close to zero, but not exactly to zero.

 

[pka]

Well... I thought (maybe wrongly) that h represents a limit, right? So in theory h will get VERY close to zero, but not exactly to zero.

You are completely correct about that. In the notation $\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$, $h\approx 0$ BUT $\bf h\ne 0$.
However I don't recall that your question had a limit in it.

 

[Steven G]

Just as pka stated, your problem did not have a limit. I will never deviate from my answer that 2h/h is not 2. I will agree that as h-->0 that 2h/h -->2

 

[JeffM]

I could not disagree more with what Jomo said here. He is allowing the tail to wag the dog.

[MATH]h \ne 0 \implies \dfrac{2h}{h} = 2[/MATH]
ALWAYS, FOREVER AND EVER, AMEN.

For example [MATH]\dfrac{6}{3} = \dfrac{2 * 3}{3} = 2.[/MATH]
The difference quotient is defined as [MATH]\dfrac{f(x + h) - f(x)}{h} \text { when } h \ne 0.[/MATH]
The derivative in standard analysis is defined as the limit of the difference quotient as h approaches zero.

Rather obviously, the value of 2 is not affected by the value of h. So the difference quotient and the derivative are exactly equal in this case although that is not the general rule.

These are definitions. No understanding is necessary to follow a definition. Understanding is needed for the meaning of the derivative. But this exercise is just asking you to do a computation according to a rule. And there is no problem of dividing by zero in calculating a difference quotient because the denominator is never zero by definition

 

[JeffM]

To elaborate on my previous post.

[MATH]g(x) = 2x^2 - 3x + 7.[/MATH]
The difference quotient assumes h is not zero and equals

[MATH]\dfrac{\{2(x + h)^2 - 3(x + h) + 7\} - (2x^2 - 3x + 7)}{h} =[/MATH]
[MATH]\dfrac{2(x^2 + 2hx + h^2) - \cancel {3x} - 3h + \cancel 7 - 2x^2 + \cancel {3x} - \cancel 7}{h} =[/MATH]
[MATH]\dfrac{\cancel {2x^2} + 4xh + 2h^2 - 3h - \cancel {2x^2}}{h} = \dfrac{4xh + 2h^2 - 3h}{h} = 4x - 3 + 2h.[/MATH]
Simple algebra.

And it is intuitive that

[MATH]\lim_{h \rightarrow 0} 4x - 3 + 2h = 4x - 3.[/MATH]
Here the difference quotient and derivative differ (as usually is the case), but there is nothing conceptually difficult about either computation.

 

[Steven G]

[MATH]h \ne 0 \implies \dfrac{2h}{h} = 2[/MATH]
ALWAYS, FOREVER AND EVER, AMEN

I agree with what you wrote.

 

[Steven G]

The difference quotient is defined as [MATH]\dfrac{f(x + h) - f(x)}{h} \text { when } h \ne 0.[/MATH]

I don't recall that being the definition of the difference quotient.

I have no problem with you saying that if from the difference quotient you get to 2h/2 that you call that fraction 2. I am sure that you are not the only one who would say that. I just am not one of them. I have absolutely nothing against your viewpoint.

 

[nosit]

Thank you all. Really appreciate, all answers were really helpful.

 

[JeffM]

@Jomo

I am not being argumentative. I am genuinely curious. How do you define the difference quotient? Does it not require the denominator to be non-zero just as all division permissible in the real numbers requires a non-zero denominator?

 

[HallsofIvy]

@Jomo

I am not being argumentative. I am genuinely curious. How do you define the difference quotient? Does it not require the denominator to be non-zero just as all division permissible in the real numbers requires a non-zero denominator?

Yes, it does. The difference quotient, $\displaystyle \frac{f(x+h)- f(x)}{h}$ requires that $\displaystyle h\ne 0$. However, the derivative is NOT the difference quotient, it is the limit of the difference quotient as h goes to 0.

I once made the mistake of agreeing to teach a course titled "Calculus for Economics and Business Administration" in the Business Administration Department. The only prerequisite for the course was basic algebra and I had to use the textbook assigned by the Business Admimistration Department. It was awful! One one page they gave the basic rules for limits:
"The limit of f+ g is the limit of f plus the limit of g."
"The limit of f- g is the limit of f minus the limit of g."
"The limit of fg is the limit of f times the limit of g."
"The limit of f/g is the limit of f divided by the limit .of g provided the limit of g is not 0."

The very next page defines the derivative as "$\displaystyle \lim_{h\to 0}\frac{f(x+h)- f(x)}{h}$" completely ignoring the fact the rule for f/g does not apply!

(If you are wondering, we need one more rule: if for some interval (a, b) that includes $\displaystyle x_0$ f(x)= g(x) for all x except $\displaystyle x_0$ (a "punctured interval") then \(\displaystyle \lim_{x\to x_0} f(x)= \lim_{x\to x_0) g(x)\).)

 

[JeffM]

Yes, it does. The difference quotient, $\displaystyle \frac{f(x+h)- f(x)}{h}$ requires that $\displaystyle h\ne 0$. However, the derivative is NOT the difference quotient, it is the limit of the difference quotient as h goes to 0.

I once made the mistake of agreeing to teach a course titled "Calculus for Economics and Business Administration" in the Business Administration Department. The only prerequisite for the course was basic algebra and I had to use the textbook assigned by the Business Admimistration Department. It was awful! One one page they gave the basic rules for limits:
"The limit of f+ g is the limit of f plus the limit of g."
"The limit of f- g is the limit of f minus the limit of g."
"The limit of fg is the limit of f times the limit of g."
"The limit of f/g is the limit of f divided by the limit .of g provided the limit of g is not 0."

The very next page defines the derivative as "$\displaystyle \lim_{h\to 0}\frac{f(x+h)- f(x)}{h}$" completely ignoring the fact the rule for f/g does not apply!

(If you are wondering, we need one more rule: if for some interval (a, b) that includes $\displaystyle x_0$ f(x)= g(x) for all x except $\displaystyle x_0$ (a "punctured interval") then \(\displaystyle \lim_{x\to x_0} f(x)= \lim_{x\to x_0) g(x)\).)

Thank you. If you look at my posts #9 and #10, you will see that I did distinguish between the difference quotient and the derivative. However, I had never tried to formulate a general rule for the limit of a quotient when the limit for the denominator was zero.