Question

How do you compute the instantaneous rate of change of `f(x)=sqrt(4x+1)` at the point (2,3) as a limit of the average rate of change of f(x) on the interval `[2,2+h]`?

Answer 1

Thank you for being specific about the method you want!

As you probably know, you need to find:

`lim_(hrarr0)(f(2+h)-f(2))/h=lim_(hrarr0)(sqrt(4(2+h)+1)-sqrt(4(2)+1))/h`

`=lim_(hrarr0)(sqrt(4h+9)-3)/h` (and here's where the question arises)

An attempt to evaluate the limit by substitution results in the indeterminate form `0/0`.

Learn this trick (technique) try rationalizing the numerator. (Sometimes you have to try something and just see if it helps. That's how mathematics works beyond the elementary levels.)

`(sqrt(4h+9)-3)/h= ((sqrt(4h+9)-3))/h ((sqrt(4h+9)+3))/((sqrt(4h+9)+3))`

`=((4h+9)-9)/(h(sqrt(4h+9)+3))=(4h)/(h(sqrt(4h+9)+3)` which is still indeterminate as `hrarr0`

For `h!=0`, `(sqrt(4h+9)-3)/h=4/(sqrt(4h+9)+3` So we can evaluate the limit.

`lim_(hrarr0)(sqrt(4h+9)-3)/h = lim_(hrarr0)4/(sqrt(4h+9)+3)=4/(3+3)=2/3`

.

Without all the discussion, it looks like this:

`lim_(hrarr0)(f(2+h)-f(2))/h=lim_(hrarr0)(sqrt(4(2+h)+1)-sqrt(4(2)+1))/h`

`=lim_(hrarr0)(sqrt(4h+9)-3)/h=lim_(hrarr0) ((sqrt(4h+9)-3))/h ((sqrt(4h+9)+3))/((sqrt(4h+9)+3))`

`=lim_(hrarr0)((4h+9)-9)/(h(sqrt(4h+9)+3))=lim_(hrarr0)4/(sqrt(4h+9)+3)=4/(3+3)=2/3`