How do you prove from the definition of differentiability that the function #f(x)=(2x+1)/(x-2)# is differentiable?

Question

Emilia Aguilar · Accepted Answer

#f# is differentiable at #x in RR-{2}, and, f'(x)=-5/(x-2)^2, x in RR-{2}.#

Let # f : RR rarr RR#, be a function and let #x in RR#. We say that the

function #f# is Differentiable at #x# , if, the following Limit

exists # : lim_(trarrx) (f(t)-f(x))/(t-x), where, t in RR, t!=x.#

Also, if the above Limit exists, it is called the Derivative of #f# at

#x#, and, is denoted by #f'(x).#

We have, #f(x)=(2x+1)/(x-2), x in RR-{2}#

#rArr f(t)=(2t+1)/(t-2), t in RR-{2}, tnex#

#:. f(t)-f(x)=(2t+1)/(t-2)-(2x+1)/(x-2)#

#={(2t+1)(x-2)-(2x+1)(t-2)}/((t-2)(x-2))#

#={(2tx-4t+x-2)-(2tx-4x+t-2)}/((t-2)(x-2))#

#=(-5t+5x)/((t-2)(x-2))=(-5(t-x))/((t-2)(x-2))#

Since, #t!=x,# we have, #(f(t)-f(x))/(t-x)=-5/((t-2)(x-2))#

Now, #lim_(trarrx) (f(t)-f(x))/(t-x)=lim_(trarrx) -5/((t-2)(x-2))#

#=-5/((x-2)(x-2))=-5/(x-2)^2#.

Thus, we find that, the Limit in question exists, and so, the given

fun. #f# is differentiable at #x in RR-{2}, and, f'(x)=-5/(x-2)^2, x in RR-{2}.#

Enjoy maths.!

Ezra Adams · Answer

undefined To prove that the function $ f(x) = \frac{2x + 1}{x - 2} $ is differentiable, we need to demonstrate that its derivative exists at every point in its domain.

Given the definition of differentiability, a function $ f(x) $ is differentiable at a point $ x = a $ if the following limit exists:

$$ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} $$

To prove differentiability for $ f(x) $, we'll calculate this limit for any arbitrary point $ x = a $ in the domain of $ f(x) $.

Let's start by finding the derivative of $ f(x) $ using the quotient rule:

$$ f'(x) = \frac{(x - 2)(2) - (2x + 1)(1)}{(x - 2)^2} $$

Now, simplify the expression:

$$ f'(x) = \frac{2x - 4 - 2x - 1}{(x - 2)^2} $$
$$ f'(x) = \frac{-5}{(x - 2)^2} $$

Now, let's evaluate the limit as $ h $ approaches 0:

$$ f'(a) = \lim_{h \to 0} \frac{\frac{2(a + h) - 4 - 2(a + h) - 1}{(a + h - 2)^2} - \frac{-5}{(a + h - 2)^2}}{h} $$

$$ = \lim_{h \to 0} \frac{\frac{-5}{(a + h - 2)^2}}{h} $$

$$ = \lim_{h \to 0} \frac{-5}{h(a + h - 2)^2} $$

$$ = \frac{-5}{(a - 2)^2} \lim_{h \to 0} \frac{1}{h} $$

Since the limit $ \lim_{h \to 0} \frac{1}{h} $ exists (it's $ \pm \infty $), we have:

$$ f'(a) = \frac{-5}{(a - 2)^2} \cdot \infty $$

Thus, the derivative $ f'(x) $ exists for all $ x $ in the domain of $ f(x) $, making $ f(x) $ differentiable everywhere in its domain.