a critical number is where f' is zero or undefined.
Using the product rule,
f'(x) = (6)(e^(4x)) + (6x+1)(4e^(4x)) = 2e^(4x)(12x+5)
Since e^(4x) is never zero, we just need
12x+5 = 0
x = -5/12
The function f(x)=(6x+1) e^4x has one critical number. Find it
I know how to solve this kind of equation; however, I never solved one with a binomial. How will can I factor this when working out the equation?
3 answers
critical values usually include any intercepts, any max/mins, and any points of inflection.
So it might be a good idea to look at the graph
www.wolframalpha.com/input/?i=plot+y+%3D+(6x%2B1)+e%5E(4x)+from+-1+to+1
- a bit hard to tell what happens as it gets near the x-axis, but surely it just keeps
increasing as x > 0
Let's look at the x-intercept, that is, when y = 0
(6x+1) e^(4x) = 0
so 6x+1 = 0 or e^(4x) = 0
x = -1/6 , but e^? = 0 has no solution (try some values of ± x)
confirmed again by: www.wolframalpha.com/input/?i=solve+(6x%2B1)+e%5E(4x)+%3D+0
Notice that my graph shows that drops a bit below the x-axis but then appears to
rise again, approaching the x-axis from below.
so we have to do some Calculus
f ' (x) = (6x+1)(4e^(4x)) + 6(e^(4x))
= 0 for a max/min
e^(4x) (24x+4 + 6) = 0
so e^(4x) = 0 , which has no solution OR 24x = -10, x = -5/12
so we have a minimum value when x = -5/12
I will leave it up to you to find that minimum if you have to, it is a small negative.
Are there any points of inflection? From our graph it looks like there is one
So f'(x) = e^(4x) (24x + 10)
f '' (x) = 24(e^(4x)) + (24x+10)(4(e^(4x))
= e^(4x) ( 24 + 96x + 40) = 0 at a point of inflection
again e^(4x) = 0 has no solution , but 96x = -64 gives us x = - 2/3
at a point of inflection
check my arithmetic
So it might be a good idea to look at the graph
www.wolframalpha.com/input/?i=plot+y+%3D+(6x%2B1)+e%5E(4x)+from+-1+to+1
- a bit hard to tell what happens as it gets near the x-axis, but surely it just keeps
increasing as x > 0
Let's look at the x-intercept, that is, when y = 0
(6x+1) e^(4x) = 0
so 6x+1 = 0 or e^(4x) = 0
x = -1/6 , but e^? = 0 has no solution (try some values of ± x)
confirmed again by: www.wolframalpha.com/input/?i=solve+(6x%2B1)+e%5E(4x)+%3D+0
Notice that my graph shows that drops a bit below the x-axis but then appears to
rise again, approaching the x-axis from below.
so we have to do some Calculus
f ' (x) = (6x+1)(4e^(4x)) + 6(e^(4x))
= 0 for a max/min
e^(4x) (24x+4 + 6) = 0
so e^(4x) = 0 , which has no solution OR 24x = -10, x = -5/12
so we have a minimum value when x = -5/12
I will leave it up to you to find that minimum if you have to, it is a small negative.
Are there any points of inflection? From our graph it looks like there is one
So f'(x) = e^(4x) (24x + 10)
f '' (x) = 24(e^(4x)) + (24x+10)(4(e^(4x))
= e^(4x) ( 24 + 96x + 40) = 0 at a point of inflection
again e^(4x) = 0 has no solution , but 96x = -64 gives us x = - 2/3
at a point of inflection
check my arithmetic
Thank you so much, I now understand where I went wrong at first!
2 answers