# `f(x)=4 - x - 3x^4` Find the points of inflection and discuss the concavity of the graph of the function.

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### 3 Answers

`f(x)=4-x-3x^4`

differentiating,

`f'(x)=-1-12x^3`

differentiating again,

`f''(x)=-36x^2`

Now to determine the inflection points set f''(x)=0

-36x^2=0

x=0

Now let us test the concavity of the function by taking test values of x=-2 and 2.

`f''(-2)=-36(-2)^2=-144`

`f''(2)=-36(2)^2=-144`

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` .**

Given: `f(x)=4-x-3x^4`

Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s).

`f'(x)=-12x^3-1`

`f''(x)=-36x^2=0`

`x=0`

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.