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.