Question

@oobleck (but anyone else can feel free to come and help)
In response to (jiskha.com/questions/1795315/Please-check-my-answers-and-give-correct-ones-I-tend-to-make-dumb-mistakes-so-Im)

#2 I got 6x^2(x^3 +5)
#3 no matter how I solve it I get -2(sqrt(2) -1)
#5 I'm sorry but I still can't seem to comprehend number 5
#6 I... don't know what I was thinking for that one... the answer would be (f(3)-f(3-0.1))/(0.1) then?
#8 I'm still confused on 8.. do you think you could solve it further?
#9 I'm confused about number 9?
#11 -5?
#13 so it would just be f'(c)?
#17 so -1/e?

Answer 1

https://www.jiskha.com/questions/1795315/Please-check-my-answers-and-give-correct-ones-I-tend-to-make-dumb-mistakes-so-Im

Answer 2

#2 ok
#3 my bad - you are correct
#5 me too. the nonzero values for f'(x) confuse me.
#6 yes. That is only an approximation over an interval
#8 we have
dy/dx = 2f'(g)*g'
so, using the product rule where y' = u*v
y" = 2(f"(g)*g')*g' + 2f'(g)*g" = 2f"(g)(g')^2 + 2f'(g)g"
#9 s(t) = -4sint - t/2 + 10
v(t) = -4cost - 1/2
v=0 when cost = -1/8, so t = t = 97.18°
a(t) = 4sint = 3.969
#11 (5x+10)/(x+2) = 5 when x ≠ -2
At x = -2 there is a hole, not an asymptote
So, if b ≠ -10 the numerator is nonzero divided by 0, giving an asymptote
#13 No. Just doing a straightforward definite integral
∫[0,c] f"(x) dx = f'(c) - f'(0)
#17 ok

Answer 3

Thank you so much for your help! It's a shame we couldn't figure out #5 tho v.v

Answer 4

Do you think were over complicating #5? If the prime of 1 is 1/2 and the prime of 2 is 3, doesn't that mean the prime of 1 1/2 has to be between 1/2 and 3. If we follow this kind of mentality only answer choice 2 satisfies it.

Answer 5

Yeah, I thought of that, and you're probably right.

You should talk to your teacher and ask what they meant by "critical" values. When we're talking calculus, that has a certain special meaning, not just "values of f'(x)"