How do you find f'(x) using the limit definition given # f(x)=3x^(−2)#?

Question

Paisley Johnson · Accepted Answer

I found #f'(x)=-6/x^3# Using the limit definition we get: #f'(x)=lim_(h->0)(f(x+h)-f(x))/h# where #h# is a small increment. In our case we have (using the fact that #x^-2=1/x^2#): #f'(x)=lim_(h->0)(3/(x+h)^2-3/(x^2))/h=# #=lim_(h->0)1/h((3x^2-3(x+h)^2)/((x+h)^2x^2))=# #=lim_(h->0)1/h((cancel(3x^2)cancel(-3x^2)-6xh-3h^2)/((x+h)^2x^2))=# #=lim_(h->0)1/cancel(h)(cancel(h)(-6x-3h)/((x+h)^2x^2))=# as #h->0# we get: #=lim_(h->0)((-6xcancel(-3h)^0)/((x+cancel(h)^0)^2x^2))=(-6x)/x^4=-6/x^3#

Isabella Ackerman · Answer

#= - 6/x^3#

by definition

#f'(x) =lim_{h to 0} (f(x+h) - f(x))/(h) #

#=lim_{h to 0} (1/h) (3/(x+h)^2 - 3/x^2)#

#=lim_{h to 0} (1/h)* (3x^2 - 3(x+h)^2)/((x+h)^2x^2)#

#=lim_{h to 0} (1/h) * (3x^2 - 3(x^2+2xh+h^2))/((x+h)^2x^2)#

#=lim_{h to 0} (1/h)* ( - 6xh-3h^2)/((x+h)^2x^2)#

#=lim_{h to 0}( - 6x-3h)/((x+h)^2x^2)#

#=lim_{h to 0}( - 6x)/((x+h)^2x^2) + mathcal(O)(h)#

#=-( 6x)/((x+0)^2x^2)#

#=-( 6x)/x^4#

#= - 6/x^3#

Ezra Adams · Answer

undefined To find $ f'(x) $ using the limit definition, we use the formula:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Given $ f(x) = 3x^{-2} $, substitute into the formula:

$$ f'(x) = \lim_{h \to 0} \frac{3(x + h)^{-2} - 3x^{-2}}{h} $$

Simplify the expression:

$$ f'(x) = \lim_{h \to 0} \frac{3}{h} \left( \frac{1}{(x + h)^2} - \frac{1}{x^2} \right) $$

$$ = \lim_{h \to 0} \frac{3}{h} \left( \frac{x^2 - (x + h)^2}{x^2(x + h)^2} \right) $$

$$ = \lim_{h \to 0} \frac{3}{h} \left( \frac{x^2 - (x^2 + 2xh + h^2)}{x^2(x + h)^2} \right) $$

$$ = \lim_{h \to 0} \frac{3}{h} \left( \frac{-2xh - h^2}{x^2(x + h)^2} \right) $$

$$ = \lim_{h \to 0} \frac{3(-2x - h)}{x^2(x + h)^2} $$

Now, we can substitute $ h = 0 $ into the expression:

$$ f'(x) = \frac{3(-2x)}{x^2 \cdot x^2} $$

$$ = \frac{-6}{x^3} $$