difference between derivative and 2 point formula

[crazyb]

Have I understood correctly that the slope could be found either by finding the output of the derivative for that specific value or by using the 2 point difference formula?

I just came across a function these days and got different results for the same value.

Say f(x)=x²-4x

Thus f(-1)=5 and f(0)=0 and so on.

the derivative would be f'(x)=2x-4

so between x=0 and x=-1 the slope would be y₂ - y_{1 /} x₂ - x<sub>1
namely 0-5/0-(-1) = -5/1=-5

However using the derivative f'(-1) would return -6

Why do I get different results? Wouldn't it be normal to add the value of the derivative f'(x)=-6 to the value of the function namely f(-1)=5 and get the value that the function returns for the next unit, that is f(0)=0. I mean 5 + (-6) = -1. It's pretty close bot not sharp 0. Why?</sub>

 

[crazyb]

[FONT=&quot] Have I understood correctly that the slope could be found either by finding the output of the derivative for that specific value or by using the 2 point difference formula?

I just came across a function these days and got different results for the same value.

Say f(x)=x²-4x

Thus f(-1)=5 and f(0)=0 and so on.

the derivative would be f'(x)=2x-4

so between x=0 and x=-1 the slope would be y₂ - y_{1 /} x₂ - x₁

namely 0-5/0-(-1) = -5/1=-5

However using the derivative f'(-1) would return -6

Why do I get different results? Wouldn't it be normal to add the value of the derivative f'(x)=-6 to the value of the function namely f(-1)=5 and get the value that the function returns for the next unit, that is f(0)=0. I mean 5 + (-6) = -1. It's pretty close bot not sharp 0. Why?

[/FONT]

 

[stapel]

Have I understood correctly that the slope could be found either by finding the output of the derivative for that specific value or by using the 2 point difference formula?

If you're referring, in the latter case, to using secant lines to find the tangent line, your course (the instructor or the textbook) should have explained that secant lines can, if chosen appropriately, approximate the tangent line's slope but, in general, they will not give the same value.

I just came across a function these days and got different results for the same value.

Say f(x)=x²-4x

Thus f(-1)=5 and f(0)=0 and so on.

the derivative would be f'(x)=2x-4

Which means that the value of the slope at x = -1 will be f'(-1) = 2(-1) - 4 = -6. But the original line, f(x) = x² - 4x, is a curved line, so the slope on either side of x = -1 will be different. And the "average slope", being a two-point secant-line approximation, will almost certainly (and almost always) have a different value. Draw sime tangents and secants on a parabola to visualize why this is so.

so between x=0 and x=-1 the slope would be y₂ - y_{1 /} x₂ - x₁

namely 0-5/0-(-1) = -5/1=-5

This is the slope of the straight line between two points which happen also to lie on the curved line. But the further you get from x = -1 (such as at x = 0), the further the secant-line's slope will wander from the tangent-line's slope. If you want to approximate with secant lines, you'll need to use very close values, like x = -1.001 and x = -0.999.

 

[Harry_the_cat]

I'd suggest you draw a graph and look at what you are trying to do graphically. It should make sense then why the "answers" are different.