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).

Asked on by cspanutius

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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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.

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