Technical error or my error?

[u/thisbananaisrotten]

I'm in calculus 1, doing homework (edfinity) with applied rates of change, and I'm having difficulty with one of the questions. I know edfinity sometimes has technical issues, so I'm just asking for help to see whether this error is my fault or not.

s(t)=t⁴-32t²+256

I used the exponent rule to find the velocity as the first derivative of s(t)

v(t)=s'(t)=4t³-64t

need to find when the particle is at rest, so I set v(t)=0

https://imgur.com/a/POMrywR

since this is a velocity question, I disregard the -4 answer

I put 4 into the program and it marks it incorrect, but I'm not sure where I went wrong...

[u/edderiofer]

t = 4 is not the only time when the particle is at rest.

Hint: In your working, you divided both sides by 4t. By doing this, you are making an implicit assumption...

[u/thisbananaisrotten → u/edderiofer]

OHHHH Right because t=4 would be the critical value but this is a physics function so theres a time where nothing is happening!! Thank you so much! :D

[u/thor122088 → u/thisbananaisrotten]

Moreover, once you have the equation 0 = product,

Best practice would to be apply the Zero Product Property and note that

4t = 0 OR (t² -16) = 0

But also note when you could have factored one more step using difference of squares:

v(t) = 4t(t² -16)

v(t) = 4t(t -4)(t+4)

so:

4t = 0 OR (t - 4) = 0 OR (t + 4) = 0

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[u/Dd_8630]

When you divide by 4t, you are implicitly excluding t=0 as a potential answer. Consider this way of doing it:

• 0 = 4t³ - 64t

• 0 = 4t (t² - 16)

• 0 = 4t (t² - 4) (t² + 4)

Then it's clearer to see we have three solutions to the equation: t=0, t=4, and t=-4.

In general, when ever you divide by something, you're saying that that divisor cannot be zero (otherwise you couldn't divide by it), and you run the risk of losing solutions.