For ethical reasons, I won't ask the exact assignment question:
For a function such as $4x^3 + 2y^3 -yx^2 = 49$. How could I use the derivative to estimate the values on the curve near a certain point? I am asked to create a table of values close to a point given to me. Since I have created the function above, I cannot give one as there is likely no pretty answer. So, if the solution requires this, how would I utilize it?
The derivative of this function is similar in nature to the one in my assignment, that being it has values of x and y. The derivative for this particular function is $$ f'(x)=\frac{2x(6x-y)}{x^2 -6y^2} $$
I believe the solution lies in the using something similar to f(x) ≈ f(a) + f'(a)(x − a), but I do not know how to implement this. For my assignment in particular it asks me, after telling me to find values on the curve near x=1, y=2, to include the 0.96, 0.98, 1, 1.02, and 1.04 in my table of values. I can only assume this means the x values and find the subsequent y? I am not sure how to do this with the derivative considering if I put in a value of x, there are two unknowns: the value of y and the derivative itself.
Sorry I have waffled a bit, and I apologise if my wording makes things confusing. It is a manifestation of my own confusion.
Edit: The next part of the question involves finding x values such that the tangent to the curve is horizontal, or vertical. How would I go about this? My initial response. for horizontal, is to make the derivative =0, but I have two variables? How could I find a vertical tangent?
If all you have learned about is the linear approximation $$ f(x)\approx f(a)+f'(a)(x-a) $$ you should simply plug in $a=1,f(a)=2$ to this formula to have the linear approximation $$ f(x)\approx 2-\tfrac8{23}(x-1) $$ which should then be ready for you to estimate $f(x)$ for $x\in\{0.96,0.98,1.02,1.04\}$.