# Let a be a positive number. The function f(x) = a x^3+4x is concave up on the intervalLet a be a positive number. The function f(x) = a x^3+4x is concave up on the interval (A) (root ([-4]/a), root...

Let a be a positive number. The function f(x) = a x^3+4x is concave up on the interval

Let a be a positive number. The function f(x) = a x^3+4x is concave up on the interval

(A) (root ([-4]/a), root (4/a) )

(B) (root ([-4]/a), 0)

(C) (-infinity , 0)

(D) (0, infinity).

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`f(x) = ax^3+4x`

`= x(ax^2+4)`

x=0

or

`ax^2+4 = 0`

`x^2 = -4/a`

since a is a positive number -4/a is negative. So there is no real roots for that expression.

`f'(x) = 3ax^2+4`

For maximum and minimum f'(x) = 0

`3ax^2+4 =0 `

`x^2 = -4/(3a)`

Since a is positive there is no x value that f'(x)=0. So there is no maximums and minimums in f(x).

`f(x) = ax^3+4x = x^3(a+4/x^2)`

`lim_(xrarr-oo) f(x) = -oo`

`lim_(xrarr oo) f(x) = oo`

- So the graph is going `-oo` to `oo` .

- It cuts x-axis only at x=0
- There is no maximum and minimum points in the graph.

**Answer (C) and (D) are correct.**

Note

The rate of increase in the graph is always increasing in the top part of the graph. So answer (D) is a best fit. But we cannot say (C) is wrong.

If a=1 the graph is shown below.