Help with calculus problem with logarithm base e!

linapril

New member
I'm at one of those annoying stages where you know what the answer is, but you just can't seem to prove it... Would really appreciate some help with this one!

The question is:
What are the intervals where function f(x)=e2x-2ex is concave and convex respectively.

I have derived f(x) to get
f'(x)=2e2x-2ex which I derived again to get the second derative:
f''(x) = 4e2x-2ex

After which I put up the equation
f''(x) = 0, since this determines the inflection point
which becomes 4e2x-2ex=0

This is where I get stuck. I know have to find the value of x, and then test values bigger and smaller of it using the equations for concave/convex parabolas, and I'm pretty sure that the answer is x=ln 0.5 but I just don't seem to be able to show it.. Logarithms was never my thing it appears.

Either way, I tried doing this
2(2e2x-ex)=0
2e2x-ex=0
2e2x=ex
2x ln 2e = x ln e

And this is where I don't know how to continue..

MarkFL

Staff member
Re: Help with calculus problem with logarithm base e!!!

You have correctly differentiated to get the second derivative:

$$\displaystyle f''(x)=4e^{2x}-2e^x$$

You factored, but you didn't factor completely. Consider:

$$\displaystyle f''(x)=2e^x\left(2e^x-1 \right)$$

Now equate that to zero to determine the critical value. which you are correct about, but you can show it using the above form. What do you find?

linapril

New member
Re: Help with calculus problem with logarithm base e!!!

I'm not quite sure I understand... Why does 2ex times 2ex equal 4e2x?
And also, If I weren't to factorise I would continue with
4e2x-2ex=0
4e2x=2ex
Is it incorrect to simplify this to
2e2x=ex
And if so, why?

MarkFL

Staff member
Re: Help with calculus problem with logarithm base e!!!

I'm not quite sure I understand... Why does 2ex times 2ex equal 4e2x?
And also, If I weren't to factorise I would continue with
4e2x-2ex=0
4e2x=2ex
Is it incorrect to simplify this to
2e2x=ex
And if so, why?
Recall the property of exponents:

$$\displaystyle a^b\cdot a^c=a^{b+c}$$

You could continue as you did to get:

$$\displaystyle 4e^{2x}=2e^x$$

Now, what happens if you multiply through by $$\displaystyle \frac{e^{-x}}{2}$$? Are you in danger of multiplying through by zero in doing so? What does the above property of exponents yield?

Deveno

Well-known member
MHB Math Scholar
I'm at one of those annoying stages where you know what the answer is, but you just can't seem to prove it... Would really appreciate some help with this one!

The question is:
What are the intervals where function f(x)=e2x-2ex is concave and convex respectively.

I have derived f(x) to get
f'(x)=2e2x-2ex which I derived again to get the second derative:
f''(x) = 4e2x-2ex

After which I put up the equation
f''(x) = 0, since this determines the inflection point
which becomes 4e2x-2ex=0

This is where I get stuck. I know have to find the value of x, and then test values bigger and smaller of it using the equations for concave/convex parabolas, and I'm pretty sure that the answer is x=ln 0.5 but I just don't seem to be able to show it.. Logarithms was never my thing it appears.

Either way, I tried doing this
2(2e2x-ex)=0
2e2x-ex=0
2e2x=ex
2x ln 2e = x ln e

And this is where I don't know how to continue..
I think you have a mistake in going from this line:

$2e^{2x} = e^x$

to this line:

$2x \ln(2e) = x\ln(e)$

If we take logarithms of both sides, we get:

$\ln(2e^{2x}) = \ln(e^x)$

the right-hand side is $x$ since $\ln$ and $e^{(\_)}$ are inverse functions.

The left-hand side is:

$\ln(2\cdot e^{2x}) = \ln(2) + \ln(e^{2x}) = \ln(2) + 2x$

$\ln(2) + 2x = x$
Which should be easy to solve for $x$.