How do i find the critical points of 3^x sinx?

  x = n·π -arctan(1/ln(3)) . . . . n = 0, 1, 2

Step-by-step explanation:

Critical points are where the derivative is zero or undefined. This function has no places where the derivative is undefined, so you need to find the values of x where the derivative is zero.

The product rule can be used to find the derivative of the function.

  y = (3^x)sin(x)

  y' = ln(3)(3^x)sin(x) +(3^x)cos(x) = (3^x)(ln(3)sin(x) +cos(x))

The factor 3^x is positive everywhere, so you will find y'=0 where ...

  ln(3)sin(x) +cos(x) = 0

Dividing by cos(x) gives ...

  ln(3)tan(x) +1 = 0

  tan(x) = -1/ln(3)

  x = -arctan(1/ln(3))

The tangent function is periodic with period π, so your solutions will have multiples of π added to this value of x:

  x = n·π - arctan(1/ln(3)) . . . . . n = 0, 1, 2

if x decreases by 4 then y increases by 10

step-by-step explanation:

i belive those are the ones that are under 90 degreese.