Rate of change question

[jgarcia]

I am not sure if I did this correctly:

The question is find the rate of change dy/dx where x=x₀:

y=1/2-x; x₀=-3

Here is my work:

y=1/2-x
y'=1-x-¹/2
y'=1+x-²/2
y'=1/(2-x)²

because x₀=-3, I plug this in into the problem...

y'=1/(2-(-3))²
y'=1/25

so ... dy/dx = (1/25)/-3

This doesn't look right.

 

[stapel]

find the rate of change dy/dx where x=x₀:

y=1/2-x; x₀=-3

Here is my work:

y=1/2-x
y'=1-x-¹/2...

How did you get the last line above? Also, what exactly is the function? As formatted, the function is \(\displaystyle y\, =\, \frac{1}{2}\, -\, x\), but you seem to be working with something else. I can't quite tell what that "something else" might be, though... :shock:

 

[jgarcia]

I assume its f(x)=y=1/2-x, where x=x₀=-3.

 

[stapel]

I assume its f(x)=y=1/2-x....

That's exactly the same thing you posted before. Does that mean that I guessed your meaning correctly, so the function is \(\displaystyle y\, =\, \frac{1}{2}\, -\, x\)? If so, then the derivative is just \(\displaystyle y'\, =\, -1\)

 

[jgarcia]

That's exactly the same thing you posted before. Does that mean that I guessed your meaning correctly, so the function is \(\displaystyle y\, =\, \frac{1}{2}\, -\, x\)? If so, then the derivative is just \(\displaystyle y'\, =\, -1\)

Thank you! I see how it makes sense now.

 

[stapel]

The answer would, of course, by very different had the function been y = 1/(2 - x) = \(\displaystyle \frac{1}{2\, -\, x}\).