The function f(x) = (9 x+2)e^{-6 x} has one critical number. Please help me find the critical number!
I read that as
f(x) = (9x + 2) e^(-6x)
f ' (x) = (9x + 2)( e^(-6x) )(-6) + 9 e^(-6x)
= 0 for max/min (critical values)
e^(-6x) [ -6(9x + 2) + 9] = 0
e^(-6x) = 0 ---> no solution
or
-54x -12 + 9 = 0
x = -1/18
if you include the x-intercepts in your critical values
then 9x+2=0
x = -2/9
y-intercept, let x = 0
y = 2 e^0 = 2
Wolfram agrees with me
http://www.wolframalpha.com/input/?i=plot++%289x%2B2%29e%5E%28-6+x%29
To find the critical number of a function, you need to find the value of x where the derivative of the function is equal to zero or undefined.
Let's start by finding the derivative of the function f(x).
f(x) = (9x + 2)e^(-6x)
To find the derivative, you can use the product rule:
The product rule states that if you have two functions u(x) and v(x), then the derivative of the product u(x) * v(x) is given by:
(u(x) * v(x))' = u'(x) * v(x) + u(x) * v'(x)
Let's apply the product rule to find the derivative of f(x):
f'(x) = ((9x + 2)' * e^(-6x)) + ((9x + 2) * (e^(-6x))')
Taking the derivative of (9x + 2) and e^(-6x):
(9x + 2)' = 9
(e^(-6x))' = -6e^(-6x)
Now substitute the derivative back into the equation:
f'(x) = (9 * e^(-6x)) + ((9x + 2) * (-6e^(-6x)))
Simplifying further:
f'(x) = 9e^(-6x) - 6(9x + 2)e^(-6x)
Now we need to find the critical number, which is the value of x where the derivative is equal to zero.
Set f'(x) equal to zero and solve for x:
9e^(-6x) - 6(9x + 2)e^(-6x) = 0
Factoring out e^(-6x):
e^(-6x)(9 - 6(9x + 2)) = 0
Simplifying:
e^(-6x)(9 - 54x - 12) = 0
e^(-6x)(-54x - 3) = 0
Since e^(-6x) is never equal to zero for any real value of x, we can conclude that the critical number occurs when the expression (-54x - 3) is equal to zero.
-54x - 3 = 0
Solving for x, we get:
-54x = 3
x = -3/54
x = -1/18
So the critical number of the function f(x) = (9x + 2)e^(-6x) is x = -1/18.