Given the function $f(x) = e^x - ax$ solve for $a$ such that $f(x) \geq 1$.

Question

So I have the function $f: \mathbb{R} \to \mathbb{R}$, $f(x) = e^x - ax$ and it is known that $a > 0$. I need to find $a$ such that $f(x) \geq 1$, $\forall x\in \mathbb{R}$.

What I have done so far is to set the derivative equal to $0$ in hopes of a minimum. So I found:

$f'(x) = e^x - a$

Set it equal to $0$ and found that the point $x = ln(a)$ is the global minimum point.

Logic led me to think that since I have to find an $a$ for which the function is $\geq 0$ for all values of $x$, all I have to do is find the $a$ for which the function is $\geq 0$ at that minimum point. Since the minimum point is $x = ln(a)$, I have to solve:

$f(ln(a)) \geq 1$

That gives me

$a - a*ln(a) \geq 1$, or

$a(1 - ln(a)) \geq 1$.

Here is where I got stuck. Is my reasoning correct? Should I have done something differently? Is there a better way to do this? And if what I have done so far is correct, how could I go about solving for $a$?

Answer 1

Hint Set $$g(x)=x-x \ln(x) $$ then $$g'(x)=-\ln(x)$$ gives you that $g(x)$ has an absolute max at $x=1$.

Since $g(1)=1$ you get that $$g(x) \leq 1 \forall x$$ with equality if and only if $x=1$.

Answer 2

Your reasoning is correct. It may help to graph the function $x(1-\ln(x))$ to see where it is at least 1. You will see a very clear critical point which should give you the solution to the problem graphically. You can then prove that this point is the only value for $a$ giving $f(x) \geq 1$.

Now note that $a(1-\ln(a))$ is our minimum value of $f$ for any given $a$. We can treat this as a function and take its derivative wrt $a$. We get $a(-1/a)+(1-\ln(a))=-\ln(a)$. Setting this equal to zero gives $a=1$. This is a maximum by the second derivative test. Therefore the largest our minimum value for $f$ could be is its value at $a(1-\ln(a))$ where $a=1$, which is $1$. Since this is a maximum, all other minimum values for $f$ are less than 1, while when $a=1$ the minimum value is 1.

Note this does not work for nonpositive $a$ since that is outside the domain of ln. Can you find another argument which works for nonpositive $a$?

Answer 3

If you let $a=e^{1-b}$, then the inequality $a(1-\ln a)\ge1$ becomes $b\ge e^{b-1}$. But $g(b)=e^{b-1}-b$ has a global minimum when $g'(b)=e^{b-1}-1=0$, i.e., at $b=1$, so $e^{b-1}\gt b$ except at $b=1$. So $a=e^{1-1}=1$ is the only value for which $e^x-ax\ge1$ for all $x$.

Answer 4

No matter what $a$ is, you'll have $f(0)=e^0-a\cdot 0=1$.

So your only hope of getting $f(x)\ge 1$ everywhere is if you have $f'(0)=0$. It turns out that there is exactly one $a$ that achieves this, namely $a=1$.

Since $e^x-ax$ is easily seen to be convex no matter what $a$ is, $f'(0)=0$ will also guarantee that it has its minimum at $x=0$.

Answer 5

HINT.-$f(0)=1$ and $f'(0)=1-a$. Since $f$ is obviously increasing from a certain value of $x\gt0$, say $x_0,$ if $1-a\lt0$ then the function $f$ is decreasing at $x=0$ so it take values less that $1$ between $0$ and $x_0$. Similarly if $1-a\gt0$ the function $f$ is increasing at $0$ so it take values less than $1$ at some domain to the left of $x=0$. Therefore we should have $1-a=0$. In other words$f$ take a mímimum at $x=0$ and we have see that this minimum is equal to $1$.