Let #f(x) = (2x)/(x-1)#. What is #f(a), f(a+h)# and the quantity #(f(a+h) - f(a))/h#, where #h# is not 0?

Answer 1

Julia Adams

#(f(a+h)-f(a))/h = -2/((a+h-1)(a-1))#

Given:

#f(x) = (2x)/(x-1)#

Then:

#f(a) = (2a)/(a-1)#

#f(a+h) = (2(a+h))/(a+h-1)#

#(f(a+h)-f(a))/h = ((2(a+h))/(a+h-1)-(2a)/(a-1))/h#

#color(white)((f(a+h)-f(a))/h) = (2(a+h)(a-1)-2a(a+h-1))/(h(a+h-1)(a-1))#

#color(white)((f(a+h)-f(a))/h) = ((color(red)(cancel(color(black)(2a^2)))-color(red)(cancel(color(black)(2a)))+color(red)(cancel(color(black)(2ha)))-2h)-(color(red)(cancel(color(black)(2a^2)))-color(red)(cancel(color(black)(2a)))+color(red)(cancel(color(black)(2ha)))))/(h(a+h-1)(a-1))#

#color(white)((f(a+h)-f(a))/h) = (-2color(red)(cancel(color(black)(h))))/(color(red)(cancel(color(black)(h)))(a+h-1)(a-1))#

#color(white)((f(a+h)-f(a))/h) = -2/((a+h-1)(a-1))#

Remarks

In this question we calculate the slope of the secant between:

#(a, f(a))#

and:

#(a+h, f(a+h))#

If we then consider the limit as #h->0#, the result gives us the instantaneous slope of #f(x)# at #x=a#

#lim_(h->0) (f(a+h)-f(a))/h = lim_(h->0) -2/((a+h-1)(a-1)) = -2/(a-1)^2#

This is also known as the derivative of #f(x)# at #x=a#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.