For the function f(x) = 1/x^2 find an expression for the slope of a tangent at the point where a = 1 using . Simplify the expression first before substitution.
2 answers
using the limits formula....
point A
f(x) = 1/(x^2)
point B
f(x+deltaX) = 1/((x+deltaX)^2)
The slope of the line between these points is:
[f(x+deltaX) - f(x)]/[(x+deltaX - x]
The slope of the tangent is given when deltaX->0.
to get you started...
[(1/(x+deltaX)^2) - 1/x^2]/[x+deltaX - x]
Get rid of the fractions in the numerator and reduce the fraction as much as possible. Then take the limit as deltaX->0.
f(x) = 1/(x^2)
point B
f(x+deltaX) = 1/((x+deltaX)^2)
The slope of the line between these points is:
[f(x+deltaX) - f(x)]/[(x+deltaX - x]
The slope of the tangent is given when deltaX->0.
to get you started...
[(1/(x+deltaX)^2) - 1/x^2]/[x+deltaX - x]
Get rid of the fractions in the numerator and reduce the fraction as much as possible. Then take the limit as deltaX->0.