Differentation

Felice

2012-05-09 1:36 am

1. Given that f(x) = x^3+3kx+5, prove that
f(x) has no turning points if k>0
f(x) has a max point and min point if k<0
2. Show that the curve y= x^2+ k/x cannot have a max point for any value of k.
3. C(t)= 20t^2e^(-3t)
Find the interval on which C'(t) >0

更新1:

3. C(t)= (20t^2)(e^(-3t)) Find the interval on which C'(t) >0

回答 (1)

myisland8132

2012-05-09 5:33 am

✔ 最佳答案

1 f(x) = x^3+3kx+5, f'(x) = 3x^2+3k

if k > 0 , f'(x) > 0 for all value of x

So, f(x) has no turning points if k>0if k < 0, f'(x) = 0 =>x = √ (-k) or -√(-k)

f''(x) = 6xSo, f(x) has a max point at x = -√(-k) and min point at x = √ (-k)2 y= x^2+ k/x, y' = 2x - k/x^2

y' = 0 => 2x^3 = k => x = (k/2)^(1/3)y'' = 2 + 2k/x^3. y''((k/2)^(1/3)) = 6 > 0

So, y= x^2+ k/x cannot have a max point for any value of k.3 C(t)= (20t^2)(e^(-3t))

C'(t) = 40texp(-3t) - 60t^2 exp(-3t) = exp(-3t)[40t - 60t^2]For C'(t) > 0

exp(-3t)[40t - 60t^2] > 0

4t - 6t^2 > 0

t(6t - 4) < 0

0 < t < 2/3

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