The below solution seems to assume that 1/0 = 0

[Darkmisc]

Homework Statement

The below question seems to assume that 1/0 = 0. Have I misunderstood something?

Relevant Equations

Chain rule

Hi everyone

In the below problem, I understand that the chain rule is being used. The derivative is then equated to zero. Since the derivative is composed of dy/du and du/dx, the derivative will equal zero if either dy/du or du/dx equals zero.

However, u would be everything under the square root sign, so dy/du would be 1/(2u^0.5). If u is equated to zero, dy/du should be undefined.

Is it a mistake for the below solutions to consider dy/du=0, or have I missed something?

Thanks

 

[anuttarasammyak]

What is u that you are wondering ? I do not find the formula.

 

[Darkmisc]

What is u that you are wondering ? I do not find the formula.

 

[Orodruin]

so dy/du would be 1/(2u^0.5).

… which is non-zero so du/dx needs to be zero for dy/dx to be zero.

 

[anuttarasammyak]

$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\frac{2x-4+2e^{2x}-5e^x}{2\sqrt{u}}$$

Obviously
$$u=(x-2)^2+(e^x-\frac{5}{2})^2 >0$$

 

[malawi_glenn]

It is pretty easy to show that $u= (x-2)^2 + (e^x - \frac{5}{2})^2$ can not be zero.
Remember, this is Pythagoras theorem basically. So the only way u can be zero is if both terms are equal to zero individually. This means x₁ = 2 and x₂ = ln(5/2). But 5/2 < e so x₂ < 1. So there is no x that can make u = 0.

 

[pasmith]

The issue only occurs if $u(x) = 0$ somewhere; but then $\min\sqrt{u(x)} = 0$, and we don't need to look at the derivative of $\sqrt{u(x)}$ (at a point where it doesn't exist) to tell us that. Also, it is enough to look for a minimum of $u(x) \geq 0$, rather than $\sqrt{u(x)}$.

 

[Orodruin]

Also, it is enough to look for a minimum of u(x)≥0, rather than u(x).

Also note that this is true for any monotonically increasing function $f$ (such as the square root on the non-negative numbers).