1. Given that F(x) = x^4 6x^2 , (a) Find the relative extrema of F. (b) Use the

Question

1. Given that F(x) = x^4 6x^2 ,

(a) Find the relative extrema of F.

(b) Use the second derivative test to determine if each relative extrema is a relative minimum or relative maximum.

(c) Find the interval(s) on which F is decreasing.

2. Find the interval(s) on which the function G(x) = x^3 /3 + 11x^2/2 +30x is (i) decreasing, and (ii) increasing.

3.The function F(x) = 2 (x^3 12 x^2 has horizontal tangents at x = 0 and at x = 8. Find the maximum of F on the interval [1,1].

4.Let F(x) = (x1)^1/3*x. Then F 0 (x) = 4x3/ 3(x1)^ 2/3 .

(a) Find the horizontal tangents of F.

(b) Where is F not differentiable?

(c) Find the minimum of F on the interval [0,1].

5.Find the horizontal asymptotes of each function both toward minus infinity and toward infinity. Let’s be extra careful with part ‘c.’ Be sure to use the identity x^2 = |x| and use the fact |x| = x when x < 0. Also, remember that a horizontal asymptote is a horizontal line. So your answers must be in the form of a horizontal line. Thus y = 42 is a possible answer, but 42 isn’t.

(a) F(x) = x+1/ x1 ,

(b) F(x) = x/ 1+x^2 .

Explanation / Answer

There are multiple questions here . i am allowed to answer only 1 at a time.Please ask other as different question

a.
F(x) = x^4 - 6x^2
F'(x) = 4*x^3 - 12*x
F''(x) = 12*x^2 - 12
put F'(x) = 0
4*x^3 - 12*x = 0
x*(4*x^2 - 12) = 0
x = 0 or x = sqrt(3) or x = -sqrt(3)
Extrema of F are:
x = 0, sqrt(3) , -sqrt(3)

b.
F''(x) = 12*x^2 - 12
at x= 0
F''(x) = 12*0^2 - 12 = -12
since F''(0) is negative, x= 0 is maxima

at x= sqrt(3)
F''(x) = 12*3 - 12 = 24
since F''(0) is positive, x= sqrt(3) is minima

at x= -sqrt(3)
F''(x) = 12*3 - 12 = 24
since F''(0) is positive, x= - sqrt(3) is minima

c.
F is decreasing in (0,sqrt(3))

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