Prove (x^5)+(e^x)=0 has at most 1 real root

Question

I know to like assume that (x^5)+(e^x)=0 has two distinct roots and then find the contradiction. But am stuck.  

Answer 1

Let f(x) = x⁵+e^x

 

f(0) = 1 > 0 and f(-1) = -1 + 1/e < 0  and f(x) is continuous for all x

 

By the Intermediate Value Theorem, f(c) = 0 for some number, c, between x = -1 and x = 0.

 

So, f has at least one real root.

 

Assume that f(x) has another real root, b. So, f(b) = f(c) = 0.  

 

f is continuous on the closed interval with endpoints b and c and is differentiable on the open interval between b and c.

 

So, by Rolle's Theorem, there is a real number, d, in the open interval between b and c such that f'(d) = 0.

 

But, f'(x) = 5x⁴ + e^x > 0 for all x.  

 

  So, f'(d) ≠ 0

 

**CONTRADICTION**

 

So, our assumption that f(x) has more than 1 real root is false.

 

Therefore, f(x) has only one real root.

 

Answer 2

Well, hopefully you know the behavior of these two graphs individually.

 

y=x⁵ crosses the x axis in exactly one place.

y = e^x has the x axis as a horizontal asymptote and is therefore always positive.

 

y = x⁵ goes to negative infinity as x approaches negative infinity.

y= e^x approaches 0 as x approaches negative infinity.

 

So when you add them up, when x is let's say -10, the sum of the functions, y=x⁵ + e^x is definitely negative.

When x = 2, the function is definitely positive.

 

Since this is a continuous function, the intermediate value theorem guarantees that there is at least one value for which this function is zero.

 

Now, both of the individual functions are always increasing, so there sum is always increasing, so once it crossed the x axis once, it can never come back down to cross it a second time.

 

Hope this helped.