# `f(x) = (x^4 + x^3 + 1)/(sqrt(x^2 + x + 1))` Estimate the intervals of concavity to one decimal place by using a computer algebra system to compute and graph `f''`.

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When using a graph of f"(x), we follow concavity test:

A a section of function f(x) is concave up when **f"(x) > 0.**

This means the segment of the graph of f"(x) is **above the x-axis.**

A a section of function f(x) is concave down when **f"(x) < 0.**

This means the segment of the graph of f"(x) is **below the x-axis.**

Using computer algebra system, here is the graph of f(x) = `(x^4+x^3+1)/(sqrt(x^2+x+1))`

For its second derivative graph:

f"(x) =

The graph of f"(x) intersects the x-axis at x`~~` -0.7 and x`~~` 0.1.

These will be used as the boundary values to set the intervals of concavity.

**These are the final answers.**

**Concave up: (-`oo` ,-0.7) and (0.1, +`oo` ` ` )**

**Concave down: (-0.7, 0.1)**

In case you prefer a more accurate data, here is maximize view of the graph of f"(x).

The graph of f"(x) intersects the x-axis at x=-0.64 and x=0.03.

These will be used as the boundary values to set the intervals of concavity.

Concave up: (-`oo` ,-0.64) and (0.03, +`oo` )

Concave down: (-0.64, 0.03)