Question

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

Answer 1

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`.

Answer 2

Do you have rules for differentiation or are you using the definition? (a definition)

Important: You'll get the same answer either way, but I don't know where you are in your study of calculus.
If you are using a definition, it's probably one of the two below:

.

`f(x)=4x^3+2x-3`

Solution 1:

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`

Solution 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`