What is the equation of the line tangent to f(x)=(x^2-x^3)e^(-x+2x^2) at x=ln5?

1 Answer
Apr 17, 2016

y=751.8-467.1x (4s.f.)

Explanation:

Changing convention from f(x) to y to calculate equation of tangent: y=(x^2-x^3)e^(2x^2-x)

Let f=x^2-x^3 and u=2x^2-x and g=e^(2x^2-x)=e^u

f'=2x-3x^2
u'=4x-1
g'=(dg)/(du)(du)/dx=e^u(4x-1)=(4x-1)e^(2x^2-x)
y'=f'g+g'f=(2x-3x^2)e^(2x^2-x)+(x^2-x^3)(4x-1)e^(2x^2-x)

y'(ln5)=(2(ln5)-3(ln5)^2)e^(2(ln5)^2-ln5)+((ln5)^2-(ln5)^3)(4ln5-1)e^(2(ln5)^2-ln5)
=-467.1 (4s.f.)

y(ln5)=((ln5)^2-(ln5)^3)e^(2(ln5)^2-ln5)=-56.13 (4s.f.)

y-y_1=y'(x-x_1)
y+56.13=-467.1(x-ln5)
y=751.8-467.1x (4s.f.)

Checking on graph:
graph{y=(x^2-x^3)e^(2x^2-x) [1.5, 1.7, -57.852, -54.594]}

At x=ln5~~1.61, y~~56.1 and the gradient is very steep (step of .01 right ~~4.5 down, so the above solution is correct.