Can you clarify your understanding for parts (a) and (b)?

[nycmathguy]

Homework Statement

Find slope of the secant line.

Relevant Equations

f(x) = x^3

Slope of a Tangent Line For f (x) = x^3.

(a) Find the slope of the secant line containing the points (2, 8) and (3, 27).

(b) Find the slope of the secant line containing the points (2, 8) and (x, f (x)), where x does not equal 2.

For (a), I just have to find m, the slope using m = delta (x)/delta(y). You say?

Can someone explain part (b)?
We are given the point (x, f(x)).
I know that y = f(x). We are given f(x) = x^3.

From this piece of information, I developed the point (x, x^3). What's next?

If my understanding for parts (a) and (b) is wrong, I need someone to put my feet in the right direction. DO NOT SOLVE the problem in my stead.

Thanks

 

[fresh_42]

These are the ideas:

The slope of the secant is the ratio $\dfrac{\Delta y}{\Delta x}$ where $\Delta$ stands for a difference. Say point one has the coordinates $(x_1,y_1)$ and point two has the coordinates $(x_2,y_2).$ Then $\Delta x= x_2 - x_1$ and $\Delta y= y_2-y_1.$ The order is important: either second minus first coordinate on both, or the other way around on both.

The tangent is a secant where both points of intersection come closer and closer until there is only one touching point left. This means we make $\Delta x$ smaller and smaller, such that the slope of the tangent becomes $\displaystyle{\lim_{x_1 \to x_2}\dfrac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}}.$

 

[Mark44]

Homework Statement:: Find slope of the secant line.
Relevant Equations:: f(x) = x^3

The more relevant equation would be the formula for the slope of a line segment between two points. Your relevant equation is really part of the Homework Statement.

Slope of a Tangent Line For f (x) = x^3.

(a) Find the slope of the secant line containing the points (2, 8) and (3, 27).

(b) Find the slope of the secant line containing the points (2, 8) and (x, f (x)), where x does not equal 2.

For (a), I just have to find m, the slope using m = delta (x)/delta(y). You say?

Can someone explain part (b)?
We are given the point (x, f(x)).
I know that y = f(x). We are given f(x) = x^3.

From this piece of information, I developed the point (x, x^3). What's next?

If my understanding for parts (a) and (b) is wrong, I need someone to put my feet in the right direction. DO NOT SOLVE the problem in my stead.

Thanks

 

[SammyS]

Homework Statement:: Find slope of the secant line.
Relevant Equations:: f(x) = x^3

Slope of a Tangent Line For f (x) = x^3.

(a) Find the slope of the secant line containing the points (2, 8) and (3, 27).

(b) Find the slope of the secant line containing the points (2, 8) and (x, f (x)), where x does not equal 2.

For (a), I just have to find m, the slope using m = delta (x)/delta(y). You say?

I say ...
The slope, m, of a line which passes through points (x₁, y₁), (x₂, y₂) is m = (y₂ - y₁) / (x₂ - x₁) .

In short, m = (Δy)/(Δx) . You have that inverted.

Finish part a - showing your work - before diving into part b.

Can someone explain part (b)?
We are given the point (x, f(x)).
I know that y = f(x). We are given f(x) = x^3.

From this piece of information, I developed the point (x, x^3). What's next?

 

[nycmathguy]

I say ...
The slope, m, of a line which passes through points (x₁, y₁), (x₂, y₂) is m = (y₂ - y₁) / (x₂ - x₁) .

In short, m = (Δy)/(Δx) . You have that inverted.

Finish part a - showing your work - before diving into part b.

Obviously, I meant to say delta (y)/delta (x).

Part (a)

Let m = slope

Our two points: (2, 8) and (3, 27).

m = (27 - 8)/(3 - 1)

m = 19/1

m = 19

How is part (b) done?

 

[nycmathguy]

These are the ideas:

View attachment 284551
The slope of the secant is the ratio $\dfrac{\Delta y}{\Delta x}$ where $\Delta$ stands for a difference. Say point one has the coordinates $(x_1,y_1)$ and point two has the coordinates $(x_2,y_2).$ Then $\Delta x= x_2 - x_1$ and $\Delta y= y_2-y_1.$ The order is important: either second minus first coordinate on both, or the other way around on both.

The tangent is a secant where both points of intersection come closer and closer until there is only one touching point left. This means we make $\Delta x$ smaller and smaller, such that the slope of the tangent becomes $\displaystyle{\lim_{x_1 \to x_2}\dfrac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}}.$

Part (a)

Let m = slope

Our two points: (2, 8) and (3, 27).

m = (27 - 8)/(3 - 1)

m = 19/1

m = 19

How is part (b) done?

 

[fresh_42]

Obviously, I meant to say delta (y)/delta (x).

Part (a)

Let m = slope

Our two points: (2, 8) and (3, 27).

m = (27 - 8)/(3 - 1)

Typo. It is $m=\dfrac{27-8}{3-2}=19.$

m = 19/1

m = 19

How is part (b) done?

The exact same way. It doesn't matter which way a point is actually described. You simply use $(x,y)=(x,f(x))$ instead of the specific numbers $(3,27)$.

Imagine (draw) a straight line through the points $(10\,\text{min}\, , \,5\,\text{miles})\; , \;(30\,\text{min}\, , \,15\,\text{miles})$ and calculate the slope. What does it tell you (part (a)) and what do you get if you consider the points $(0\,\text{min}\, , \,0\,\text{miles})$ and $(x\,\text{min}\, , \,f(x)\,\text{miles})$ instead (part (b))?

 

[nycmathguy]

Typo. It is $m=\dfrac{27-8}{3-2}=19.$

The exact same way. It doesn't matter which way a point is actually described. You simply use $(x,y)=(x,f(x))$ instead of the specific numbers $(3,27)$.

Imagine (draw) a straight line through the points $(10\,\text{min}\, , \,5\,\text{miles})\; , \;(30\,\text{min}\, , \,15\,\text{miles})$ and calculate the slope. What does it tell you (part (a)) and what do you get if you consider the points $(0\,\text{min}\, , \,0\,\text{miles})$ and $(x\,\text{min}\, , \,f(x)\,\text{miles})$ instead (part (b))?

Let m = slope of the secant line.

Our points: (2, 8) and (x, f (x))

I can let f(x) = x^3.

m = (x^3 - 8)/(x - 2)

On top we have the difference of cubes.

m = (x - 2)/(x - 2)(x^2 + 2x + 4)

m = 1/(x^2 + 2x + 4)

Yes?

 

[fresh_42]

Let m = slope of the secant line.

Our points: (2, 8) and (x, f (x))

I can let f(x) = x^3.

m = (x^3 - 8)/(x - 2)

On top we have the difference of cubes.

m = (x - 2)/(x - 2)(x^2 + 2x + 4)

m = 1/(x^2 + 2x + 4)

Yes?

Yes, that is the general slope, which varies if we consider different points. We are allowed to divide by $x-2$ because $x\neq 2$ has explicitly be mentioned.

You should try my simpler example with the car to see why the slope is an interesting quantity.

 

[Mark44]

m = (x^3 - 8)/(x - 2)
On top we have the difference of cubes.

Right.

m = (x - 2)/(x - 2)(x^2 + 2x + 4)
m = 1/(x^2 + 2x + 4)

No, you have switched numerator and denominator.

Yes, that is the general slope

I don't think you read what nycmathguy wrote very carefully...

Also, (x - 2)/(x - 2)(x^2 + 2x + 4) is ambiguous.
Taken literally, this is $$\frac{x - 2}{x - 2} (x^2 + 2x + 4)$$

What you meant probably was $$\frac{x - 2}{(x - 2)(x^2 + 2x + 4)}$$

Writing the expression like this -- (x - 2)/[(x - 2)(x^2 + 2x + 4)] -- would have eliminated the ambiguity, but is incorrect for the reason mentioned above.

 

[fresh_42]

I don't think you read what nycmathguy wrote very carefully...

Yes, I was so focused on the polynomials that I didn't recognize that he switched numerator and denominator. The linear notation didn't help either.

@nycmathguy:
Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

 

[nycmathguy]

Yes, I was so focused on the polynomials that I didn't recognize that he switched numerator and denominator. The linear notation didn't help either.

@nycmathguy:
Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

I appreciate the LaTex link but I'd rather use the MathMagic Lite app. A million times easier to use and learn than LaTex.