Question 1040466
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Part A)


y = x*4^x
dy/dx = 4^x+x*ln(4)*4^x ... apply derivative; use product rule and log rule
dy/dx = 4^x * [ 1+x*ln(4) ] ... factor


Set dy/dx equal to zero. Then solve for x


4^x * [ 1+x*ln(4) ] = 0
4^x = 0 or 1+x*ln(4) = 0


4^x = 0 has no solutions


let's solve 1+x*ln(4) = 0


1+x*ln(4) = 0
x*ln(4) = -1
x = -1/ln(4)
x = -0.72134752044449


So dy/dx is only equal to zero when x = -0.72134752044449


If x is some smaller value, say x = -1, then 


dy/dx = 4^x * [ 1+x*ln(4) ] 
dy/dx = 4^(-1) * [ 1+(-1)*ln(4) ] 
dy/dx =  -0.09657359027998


So dy/dx is negative for x values such that x < -0.72134752044449


This means the original function is decreasing when x < -0.72134752044449. So on the interval (-infinity, -0.72134752044449)
Rounding to two decimal places gives <font color=red>(-infinity, -0.72)</font>


Let's plug in a value to the right of -0.72134752044449, say x = 0


dy/dx = 4^x * [ 1+x*ln(4) ] 
dy/dx = 4^(0) * [ 1+(0)*ln(4) ] 
dy/dx =  1


telling us that f(x) is increasing on the interval (-0.72134752044449, infinity)
Rounding to two decimal places gives <font color=green>(-0.72, infinity)</font>


To wrap up part a) we found that...
f(x) is decreasing on the interval <font color=red>(-infinity, -0.72)</font>
f(x) is increasing on the interval <font color=green>(-0.72, infinity)</font>
I'm using interval notation


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Part B)


In part A),we found that the function is decreasing on the interval from -infinity to -0.72134752044449. Then the function increases back again on the interval from -0.72134752044449 to infinity.


The graph will head downhill until it reaches x = -0.72134752044449. Then it will head back uphill. The question is: what is the corresponding y value for x = -0.72134752044449 ?


Let's find out by plugging in x = -0.72134752044449


y = x*4^x
y = -0.72134752044449*4^(-0.72134752044449)
y = -0.26536892271152
y = <font color=blue>-0.27</font> ... rounding to two decimal places


The smallest y can get is -0.27


Therefore, the range in interval notation is <font color=blue>[-0.27, infinity)</font>


Take note how I used a square bracket to include the endpoint. 


Here is a graph


{{{ graph( 500, 500, -5, 5, -2, 3, 0,x*4^x) }}}
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