Step by step solution to "What is the equation of the line tangent..."

Kennedy Ford

Answered question

2022-04-09

What is the equation of the line tangent to

f (x) = (x^{2} - x) e^{1 - x^{2}}

at x=1?

tabuevniru8op

Beginner2022-04-10Added 14 answers

Explanation:
Use Product Rule to find f'(x) then put in 1 for x to find the slope.
Let

f = x^{2} - x

and

g = e^{1 - x^{2}}

f'=2x-1 and

g^{'} = - 2 x e^{1 - x^{2}}

f'(x)=fg'+gf'

f^{'} (x) = (- 2 x^{3} + 2 x^{2}) e^{1 - x^{2}} + (2 x - 1) e^{1 - x^{2}}

f^{'} (1) = (- 2 + 2) e^{1 - 1} + (2 - 1) e^{1 - 1} = 0 + 1 e^{0} = 1

Equation of tangent line
y=f(a)+f'(a)(x-a)
y=f(1)+f'(1)(x-1)

f (1) = (1 - 1) e^{1 - 1} = 0

y=0+1(x-1)
y=x-1

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