Find f(a) where a is in the domain of f.
f(x) = -1/x^2
I got a few steps into it and failed :(
The equation she gave me to plug it into is
(f(a+h)-f(a))/h
It looks like you are finding the derivative by First Principles.
f(a+h) = -1/(a+h)^2
f(a) = -1/a^2
then (f(a+h)-f(a))/h
= [-1/(a+h)^2 - (-1/a^2)]/h
= [-a^2 + (a+h)^2]/[(a^2(a+h)^2](1/h)
= [-a^2 + a^2 + 2ah + h^2]/[(a^2(a+h)^2](1/h)
= h(2a+h)/[(a^2(a+h)^2](1/h)
= (2a+h)/[(a^2(a+h)^2]
Now I will assume that you are going to take the limit of that as h --> 0
Am I right?
If so, then
LIM (2a+h)/[(a^2(a+h)^2] as h --> 0
= LIM (2a)/[(a^2(a)^2]
= 2a/a^4
= 2/a^3
Thank you so much, this was very helpful :)
No worries! Let's work through it step by step.
To find f(a), we simply need to substitute 'a' into the function f(x) = -1/x^2. So, f(a) = -1/a^2.
Now, let's move on to the second part. We have the equation (f(a+h) - f(a))/h. This is an expression that represents the slope of the line between two points on the graph of the function f(x).
To evaluate this expression, we need to find f(a+h) first. We substitute (a+h) into the function f(x): f(a+h) = -1/(a+h)^2.
Now, let's substitute these values back into the equation: (f(a+h) - f(a))/h = [(-1/(a+h)^2) - (-1/a^2)] / h.
To simplify, we need to combine the fractions. Multiplying the numerator and denominator of the first fraction by a^2 and the numerator and denominator of the second fraction by (a+h)^2, we get: [-a^2/(a+h)^2 + a^2/(a^2(a+h)^2)] / h.
Now, we can combine the fractions by finding a common denominator: [(-a^2(a^2)+a^2(a+h)^2)/((a+h)^2a^2)] / h.
Simplifying further, we have: [(-a^4+a^4+2a^3h+ah^2)/((a^2+2ah+h^2)a^2)] / h.
We can cancel out the common factors in the numerator and denominator: (2a^3h+ah^2)/((a^2+2ah+h^2)a^2) / h.
Finally, we cancel out the 'h' in the numerator and denominator: (2a^3+ah)/((a^2+2ah+h^2)a^2).
This expression represents the limit of (f(a+h) - f(a))/h as h approaches 0. However, if you want to find a specific value for f(a), you can substitute a specific value for 'a' into this expression.
Keep in mind that while evaluating this expression provides you with the limit, it may not give you the exact value of f(a) for every value of 'a'.