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i am not a doctor but i will see what i can help you

y'(0) does not exist
5x + 4 = 0 when x = -4/5
so y'(-4/5) = 0
y'(-4/5) is your critical point

Well if x = 0 and x = -4/5 are you critical points, find f(0) and f(-4/5)

f(0) = 0

I found them above. I'm just having trouble numerically working out the second one.

something is amiss

@phi , Line 4, did you just multiply the left portion by 5/5?

yes, because it would be nice to have 5^3 in the bottom, so we can take its cube root.

Makes sense, just making sure. Thanks so much, dude, this helps a lot.