A function is given as f(x)=〖2xe〗^3x?

Note: Using * as multiply

a.

x = 0 , gives f(x) = 0

(0,0)

b/

f'(x)

first let u = 3x

du = 3

f(u) = (2/3)u*e^u

multiplication rule

d(rs)/du = r*ds +s*dr

so r = (2/3)u

dr = 2/3

s =e^u

ds = e^u

d(rs) = (2/3)u*e^u+ e^u(2/3)

Now use chain rule

df(u) /du = (2/3)u*e^u+ e^u(2/3)

therefore

d(f(x)) /dx = df(u)/du * du/dx = 3* ((2/3)u*e^u+ e^u(2/3) )

d(f(x)) /dx = 2*u*e^u + 2*e^u

now replace u with 3x

df(x)/dx = 2*(3x)e^(3x) +2*e^(3x) = (6x + 2)*e^3x

f'(x) = (6x + 2)*e^3x

c.

tangent at x =1

f(x) at x =1 = 2e^3

so one point is (1,2*e^3)

slope = (6*1 + 2 )*e^3 = 8*e^3

y = 8*e^(3*x) + b

plug in

(1,2*e^3)

y = 8*e^(3*1) +b

2*e^3 = 8*e^3 +b

b = -6*e^3

so tangent is

y=8*e^3*x - 6*e^3

d.

the normal has slope m_norm=-1/(8*e^3)

y = (-1/(8*e^3))*x +b

plug in point (1, 2*e^3)

2e^3 = -1/(8*e^3)*1 + b

b = +1/(8*e^3) - 2*e^3

b = (-16e^6 + 1)/2e^3

normal line

f_norm(x) = -x/(8*e^3) + (-16*e^6 +1)/(2*e^3)

e, stationary points

where the derivative is zero of undefined.

so first

f'(x) = (6x + 2)*e^3x

0 = (6x+2)

or

0 = e^3x

6x + 2 = 0

6x = -2

x = -1/3

f(-1/3) = 2*(-1/3)*e^(-1/3)*(3 ) = -2/3 *e^(-1)

f(-1/3) = -2/ (3e)

so the one stationary point is (-1/3, -2/(3*e) )

this is an absolute and relative minimum

reason why it is absolute max and relative max

the derivative is positive from values greater x =-1/3 and it rises

toward infinity (if you graph points)

The derivative is negative from x = -infinity to value less x = -1/3

so it is decreasing. Therefore this must be a relative minimum

However, by looking at the graph, you can see goes no lower and

x =+infinity it is rising to infinity

at x = -infinity is approaching 0.

e^3x never equals 0, the above is the only stationary point