Given f(x)=3x∙e^x?

f(x) = 3x e^x

a.

dy/dx = 3 d/dx(x) e^x + 3x d/dx(e^x) --- product rule

dy/dx = 3 (1) e^x + 3x e^x

= 3 e^x (1+x)

b.

Tangent at x=2 is f'(2)

f'(x) = 3e^x (1+x)

f'(2) = 3e^2(1+2)

= 9e^2

c.

Normal is the negative reciprocal of the tangent

= -1/ (9e^2)

d.

set dy/dx = 0

3 e^x (1+x)=0

1+x=0

x= -1 is a turning point

dy/dx = 3 e^x +3x e^x

d^2y/dx^2 = 3 e^x + 3 e^x + 3x e^x

= 6e^x + 3x e^x

plug in x=-1

f''(-1) = 3e^(-1) (0) + 3e^(-1) = 3/e

3/e > 0 so f(x) has a minimum at x=-1

The minimum is f(-1)

= 3(-1) e^(-1)

= -3/e

e.

see (d)

a)

y = 3x * e^x

y' = 3 * e^x + 3x * e^x

b)

y' = 3 * e² + 3 * 2 * e²

y' = 9e² --> slope (m)

y = 3x * e^x

y = 3 * 2 * e²

y = 6e²

y = mx + n

6e² = 9e² * 2 + n

6e² = 18e² + n

n = -12e²

y = 9e²x -12e² --> tangent at x=2

y = f(x) = 3xeˣ

dy/dx = f'(x) = 3eˣ + 3xeˣ

:::::

f(2) = 6e²

slope of tangent at (2,6e²) = f'(2) = 3e² + 6e² = 9e²

tangent:

y-6e² = 9e²(x-2)

y = 9e²x - 12e²

:::::

slope of normal at (2,6e²) = -1/(9e²)

normal:

y-6e² = (-1/(9e²))(x-2)

y = (-1/(9e²))x + 2/(9e²) + 54e⁴/(9e²) = (-1/(9e²))x + (2+54e²)/(9e²)

:::::

derivative = 0 at minima/maxima

3eˣ + 3xeˣ = 0

eˣ(3+3x) = 0

3+3x = 0

x = -1

f(-1) = -3/e

d²y/dx² = f"(x) = 3eˣ + (3eˣ + 3xeˣ) = 6eˣ + 3xeˣ

f"(-1) = 6/e - 3/e = 3/e > 0

(-1,-3/e) is a minimum

A