What is the slope of the line normal to the tangent line of #f(x) = e^-x+x^2-x # at # x= 0 #?

Question

What is the slope of the line normal to the tangent line of #f(x) = e^-x+x^2-x  # at # x= 0  #?

Paisley Johnson · Accepted Answer

Slope of the normal to #f(x)# at #x=0# is #1/2# #f(x) = e^(-x)+x^2-x# Differentiating: #f'(x) =-e^(-x)+2x-1# To find the slope of #f(x)# at #x=0#: #f'(0) = -1+0-1 =-2# Hence slope of the tangent #(m)# to #f(x)# at #x=0# is #-2# Since the slope of the normal at the same point = #-1/m# Therefore the slope of the normal to #f(x)# at #x=0 = -1/(-2) = 1/2#

Cameron Alexander · Answer

undefined To find the slope of the line normal to the tangent line of $f(x) = e^{-x} + x^2 - x$ at $x = 0$, we first need to find the derivative of $f(x)$ and evaluate it at $x = 0$ to find the slope of the tangent line. Then, the slope of the line normal to the tangent line will be the negative reciprocal of this slope.

The derivative of $f(x)$ is $f'(x) = -e^{-x} + 2x - 1$. Evaluating $f'(x)$ at $x = 0$, we get $f'(0) = -e^0 + 2(0) - 1 = -1$.

So, the slope of the tangent line at $x = 0$ is $-1$.

The slope of the line normal to the tangent line will be the negative reciprocal of $-1$, which is $1$. Therefore, the slope of the line normal to the tangent line of $f(x)$ at $x = 0$ is $1$.