Find constants a and b in the function f(x)=axe^bx such that f(1/5)=1 and the function has a local maximum at x=1/5.

F (x) = axe ^ bx

F ‘ (x) = d/dx (axe ^ bx)

= a (d/dx) (xe ^ bx)

= a [(d/dx * x) e ^ bx + (x * d/dx * e ^ bx)]

= a [e ^ bx + bxe ^ bx]

= a [1 + bx] e ^ bx

For critical points, set f ‘ (x) = 0 and solve for x

a [1 + bx] e ^ bx = 0

= 1 + bx = 0

= bx = -1

X = -1/b

Given that f (x) has local max at 1/5, critical point x = 1/5

-1/5 = -1/b

Therefore b = -5

F(x) = axe ^ -5x

F(1/5) = a/5 e ^-1

1 = a/5e

Therefore, a = 5e

The value of a and b is and , respectively.

Step-by-step explanation:

The given function is .

The value of function at is and the function has a local maxima at  .

So, the first derivative of the function at will be zero.

Now, calculating for a and b.

Differentiate the function,

Solving for a as,

Therefore, the value of a and b is and , respectively.

For more details, refer the link:

link

b. 18

step-by-step explanation:

so the equation for this is:

ab^2=ac⋅ad

ab being the top length of 6 and ac being the length of 2 and ad being the length of x. let's plug into the equation:

6^2 = 2x

now lets just solve for x:

36 = 2x

18 = x

i hope that !