What is the slope of the line normal to the tangent line of #f(x) = e^(x-1)+x-2 # at # x= 2 #?

Question

What is the slope of the line normal to the tangent line of #f(x) = e^(x-1)+x-2  # at # x= 2  #?

William Abdalla · Accepted Answer

The reqd. slope#=-1/(e+1)#.

We know that, for a curve #y=f(x), f'(a)# gives the slope of tgt. line to #C# at the pt.#P(a,f(a))#. As the normal to #C# at #P# is #bot# to tgt., its slope is, #-1/(f'(a))#, if, #f'(a) ne 0#. We have, #f(x)=e^(x-1)+x-2 rArr f'(x)=e^(x-1)+1# #rArr f'(2)=e+1 ne0#. Therefore, the reqd. slope#=-1/(e+1)#. Enjoy Maths.!

Isaiah Allen · Answer

undefined To find the slope of the line normal to the tangent line of $ f(x) = e^{x-1} + x - 2 $ at $ x = 2 $, we first need to find the derivative of the function $ f(x) $ and evaluate it at $ x = 2 $ to get the slope of the tangent line. Then, we can find the negative reciprocal of this slope to get the slope of the line normal to the tangent line.

First, let's find the derivative of $ f(x) $:
$$ f'(x) = \frac{d}{dx} (e^{x-1} + x - 2) $$

Using the chain rule and derivative of $ e^x $, we have:
$$ f'(x) = e^{x-1} + 1 $$

Now, let's evaluate $ f'(x) $ at $ x = 2 $:
$$ f'(2) = e^{2-1} + 1 = e + 1 $$

This is the slope of the tangent line at $ x = 2 $.

The slope of the line normal to the tangent line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the line normal to the tangent line at $ x = 2 $ is $ -\frac{1}{e + 1} $.