What is the derivative of the function and the slope at the given two points?

[onesun0000]

I was asked to find the derivative of the function [MATH]f(x)=x^3+2x[/MATH] using the limit definition of derivatives and find the slope at [MATH](1,3)[/MATH] and [MATH](-1,-3)[/MATH].

Here's my solution for the derivative of the function:


I have no idea how to find the slope at the points [MATH](1,3)[/MATH] and [MATH](-1,-3)[/MATH].

 

[pka]

I was asked to find the derivative of the function [MATH]f(x)=x^3+2x[/MATH] using the limit definition of derivatives and find the slope at [MATH](1,3)[/MATH] and [MATH](-1,-3)[/MATH].
I have no idea how to find the slope at the points [MATH](1,3)[/MATH] and [MATH](-1,-3)[/MATH].

So what? You have shown that the derivative \(f'(x)=3x^2+2\)
Do you know that the derivative is the slope of the curve?

 

[onesun0000]

So what? You have shown that the derivative \(f'(x)=3x^2+2\)
Do you know that the derivative is the slope of the curve?

Oh. It means that the slope at at points are [MATH]f'(1)=5[/MATH] and [MATH]f'(-3)=-1[/MATH]?

 

[pka]

If \(f'(x)=3x^2+2\) then \(f'(-1)=5\)

 

[Steven G]

How did you conclude that f'(-3) = -1???

You found f'(x), that is what goes between the ( ) in f( x ) is x!! Yet you plugged in a y value. Why? You do know that in a point (a, b) that the first value is a and the 2nd value is y.

 

[JeffM]

The derivative here specifies the value of the slope as a function of x. In other words, the value of y is irrelevant to the value of the slope at (x, y).

If you were asked what is the equation of tangent to the function at those points, then y would be relevant.

Do you see why y is relevant in one case and irrelevant in the other?

 

[onesun0000]

How did you conclude that f'(-3) = -1???

You found f'(x), that is what goes between the ( ) in f( x ) is x!! Yet you plugged in a y value. Why? You do know that in a point (a, b) that the first value is a and the 2nd value is y.

my mad. It's 5.

 

[lookagain]

Here's my solution for the derivative of the function:
View attachment 21946

The second to the last line needs to be just

\(\displaystyle 3x^2 \ + \ 2, \ \ \) because you've taken the limit as h approaches 0 in the last
expression from the prior line to get to the expression in the second to the last line,
which is only in terms of the variable x (and any constants).