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]?

1 Answer
Jim H
Mar 27, 2015

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

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