Now to determine the inflection points set f''(x)=0
Now let us test the concavity of the function by taking test values of x=-2 and 2.
So the function is concave down in the intervals (-`oo` ,0) and (0,`oo` ).
Since the concavity of the function is not changing , so the inflection points of the function do not exist.
You need to find the inflection points, hence, you need to solve for x the equation `f''(x) = 0` .
First, you need to determine the first derivative:
`f'(x) = (4 - x - 3x^4)' => f'(x) = -1 - 12x^3`
You need to determine now the second derivative:
`f''(x) = ( -1 - 12x^3)' => f''(x) = -36x^2`
Now, you may solve for x the equation `f''(x) = 0` .
`-36x^2 = 0 => x^2 = 0 => x_1 = x_2 = 0`
Hence, the function has two coincident inflection points at `x = 0` . You should notice that the second derivative is always negative, for all x in R, hence the function is concave downward for `x in R` .
Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s).
The critical value for the second derivative is x=0.
If f''(x)>0, the curve is concave up in the interval.
If f''(x)<0, the curve is concave down in the interval.
Choose a value for x that is less and greater than 0.
f''(-1)=-36 the graph is concave down in the interval `(-oo, 0)`
f''(1)=-36 the graph is concave down in the interval `(0,oo).`
This function is essentially always concave down.