Question

On what interval is the curve \({\rm{y = }}\int\limits_{\rm{0}}^{\rm{x}} {\frac{{{{\rm{t}}^{\rm{2}}}}}{{{{\rm{t}}^{\rm{2}}}{\rm{ + t + 2}}}}} \)concave downward?

Short answer

\({\rm{( - 4,0)}}\)

Step-by-step solution

Find the second derivative.

Consider the function

\({\rm{y = }}\int\limits_{\rm{0}}^{\rm{x}} {\frac{{{{\rm{t}}^{\rm{2}}}}}{{{{\rm{t}}^{\rm{2}}}{\rm{ + t + 2}}}}} \)

In general,

If the second derivative of the function\({{\rm{y}}^{\rm{'}}}{\rm{ = f(x)}}\)\(\frac{{{\rm{d}}{{\rm{f}}^{\rm{2}}}{\rm{(x)}}}}{{{\rm{d}}{{\rm{x}}^{\rm{2}}}}}\), is less than 0 for all\({\rm{x}}\)in the interval\({\rm{(a,b)}}\), then the graph of the function is concave downward on\({\rm{(a,b)}}\).

Need to find the second derivative of\({\rm{y}}\).

Now, use the Fundamental Theorem of Calculus PART 1, to find\({{\rm{y}}^{\rm{'}}}\).

In general, Fundamental Theorem of Calculus PART 1 shows that

If\({\rm{f}}\)is a continuous function, and

\({\rm{y = }}\int\limits_{\rm{a}}^{\rm{x}} {{\rm{f(t)dt}}} \)

Then \({{\rm{y}}^{\rm{'}}}{\rm{ = f(x)}}\)

Working on the given equation.

From this,\({{\rm{y}}^{\rm{'}}}{\rm{ = }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + x + 2}}}}\)

The second derivative of the function is the derivative of the first derivative.

Therefore,

Assigning the final value

This becomes zero when\({{\rm{x}}^{\rm{2}}}{\rm{ + 4x < 0}}\).

Need to solve this for the interval.

\(\begin{aligned}{c}{{\rm{x}}^{\rm{2}}}{\rm{ + 4x < 0}}\\{\rm{x(x + 4) < 0}}\\{\rm{x < 0}}\\{\rm{or}}\\{\rm{x + 4 < 0}}\\{\rm{x < - 4}}\end{aligned}\)

Therefore the interval on which the function is concave downward is \({\rm{( - 4,0)}}\).