For the function f(x) =1 over x squared, determine a general expression for the slope of a tangent using the limit as h approaches 0 of IROC equation

f(x+h) = 1/(x+h)^2

f(x) = 1/x^2
f(x+h) - f(x) = 1/(x+h)^2 - 1/x^2
=x^2/[x^2(x+h)^2] - (x+h)^2 /[x^2(x+h)^2]
=[ x^2 - (x+h)^2 ]/ [x^2(x+h)^2]
= [ - 2 x h - h^2 ]/[x^2(x+h)^2]
divide by h
= [ - 2 x - h ]/[x^2(x+h)^2]
let h ---> 0
=-2x/x^4
= -2/x^3

f(x)=1/x^2

f(x+d)=1/(x+d)^2

slope= lim as d>o of 1/d (1/( x+d)^2-1/x^2)=
= lim 1/d* (x^2-x^2-2dx-x^2)/(x^2*(x+d)^2) =lim (-2dx)/(x^2(x+d)^2)=-2/x^3

To determine a general expression for the slope of a tangent to the function f(x) = 1/x^2 using the limit as h approaches 0 of the Instantaneous Rate of Change (IROC) equation, we can follow these steps:

Step 1: Write down the IROC equation.
The Instantaneous Rate of Change (IROC) equation is given by:

IROC = (f(x + h) - f(x)) / h

Step 2: Substitute the function f(x) = 1/x^2 into the IROC equation.
We replace f(x) with 1/x^2:

IROC = (1 / (x + h)^2 - 1 / x^2) / h

Step 3: Simplify the numerator.
To simplify the numerator, let's find a common denominator:

IROC = ((x^2 - (x + h)^2) / (x^2(x + h)^2)) / h

Step 4: Expand and simplify the numerator further.
Expanding (x + h)^2 gives us:

IROC = ((x^2 - (x^2 + 2xh + h^2)) / (x^2(x + h)^2)) / h

Simplifying the numerator:

IROC = ((x^2 - x^2 - 2xh - h^2) / (x^2(x + h)^2)) / h

Further simplification leads to:

IROC = (-2xh - h^2) / (x^2(x + h)^2 * h)

Step 5: Cancel out h in the numerator and denominator.
After canceling out h, we get:

IROC = (-2x - h) / (x^2(x + h)^2)

Step 6: Take the limit as h approaches 0.
To find the slope of the tangent, we take the limit of the IROC equation as h approaches 0:

lim(h→0) [(-2x - h) / (x^2(x + h)^2)]

Taking the limit results in:

Slope of tangent = -2x / (x^2)^2

Simplifying further, we get:

Slope of tangent = -2 / x^3

Therefore, the general expression for the slope of a tangent to the function f(x) = 1/x^2 is -2/x^3.