How do you find the instantaneous rate of change of #y=4x^3+2x-3# at #x=2#?

Question

Isabella Ackerman · Accepted Answer

The answer is: #50#. The instantanous rate of change of the function #f(x)# in #x=a# is the slope of the tangent line to the graph of the function in that point: #m=y'(a)#. So: #y'=12x^2+2# and: #y'(2)=12*4+2=50#.

Aurora Alvarado · Answer

Are you applying the definition or do you have rules for differentiation? Important: If you are using a definition, it's probably one of the two below. You'll get the same answer either way, but I'm not sure where you are in your calculus study. . #f(x)=4x^3+2x-3#   First Solution: Using definition:  #lim_(xrarr2)(f(x)-f(2))/(x-2)# #lim_(xrarr2)(f(x)-f(2))/(x-2)=lim_(xrarr2)([4x^3+2x-3]-[4(2)^3+2(2)-3])/(x-2)# #=lim_(xrarr2)(4x^3+2x-3-33)/(x-2)=lim_(xrarr2)(4x^3+2x-36)/(x-2)# Trying to evaluate this limit by substitution gives indeterminate form: #0/0#.  But don't give up hope! Because #2# is a zero of the polynomial numerator, we can be sure that #(x-2)# is a factor. #4x^3+2x-36=(x-2)(4x^2+8x+18)# (by division or by trial and error or, perhaps, by grouping) Resuming:
#lim_(xrarr2)(f(x)-f(2))/(x-2)=lim_(xrarr2)(4x^3+2x-36)/(x-2)# #=lim_(xrarr2)((x-2)(4x^2+8x+18))/(x-2)=lim_(xrarr2)(4x^2+8x+18)# #=4(2)^2+8(2)+18=16+16+18=50# Option 2: Use the definition:  #lim_(hrarr0)(f(2+h)-f(2))/h# #lim_(hrarr0)(f(2+h)-f(2))/h# #=lim_(hrarr0)([4(2+h)^3+2(2+h)-3]-[4(2)^3+2(2)-3])/h# #=lim_(hrarr0)([4(8+12h+6h^2+h^3)+2(2+h)-3]-[33])/h# #=lim_(hrarr0)([32+48h+24h^2+4h^3+4+2h-3]-[33])/h# #=lim_(hrarr0)(50h+24h^2+4h^3)/h# #=lim_(hrarr0)(50+24h+4h^2)=50#

William Abdalla · Answer

undefined To find the instantaneous rate of change of $ y = 4x^3 + 2x - 3 $ at $ x = 2 $, you can first find the derivative of the function with respect to $ x $, which represents the rate of change of $ y $ with respect to $ x $ at any given point. Then, evaluate the derivative at $ x = 2 $ to find the instantaneous rate of change.

The derivative of the function $ y = 4x^3 + 2x - 3 $ is $ y' = 12x^2 + 2 $.

Evaluating the derivative at $ x = 2 $, we get:
$ y'(2) = 12(2)^2 + 2 $
$ y'(2) = 12(4) + 2 $
$ y'(2) = 48 + 2 $
$ y'(2) = 50 $

So, the instantaneous rate of change of $ y = 4x^3 + 2x - 3 $ at $ x = 2 $ is $ 50 $.