Chapter 5.4, Problem 23E

To determine

To find: Concave downward of the interval of the curve

Answer to Problem 23E

y is concave down when $\text{4<x<0}$ .

Explanation of Solution

Given information: the curve is $\text{y=}{\displaystyle {\int }_0^{\text{x}}\frac{{\text{t}}^2}{{\text{t}}^2\text{+t+2}}}\text{ dt}$

given, $\text{y=}{\displaystyle {\int }_0^{\text{x}}\frac{{\text{t}}^2}{{\text{t}}^2\text{+t+2}}}\text{ dt}$

Using the fundamental theorem of calculus part 1 , then we have

  $\text{<msup><mo>y</mo><mo>′</mo></msup>=}\frac{{\text{x}}^2}{{\text{x}}^2\text{+x+2}}$

Now derivative both with respect x. we have

  $\text{<msup><msup><mo>y</mo><mo>′</mo></msup><mo>′</mo></msup>=}\frac{{\text{(x}}^2\text{+x+2)}(2\text{x})-{\text{x}}^2\text{(2x+1)}}{{{\text{(x}}^2\text{+x+2)}}^2}$

  $\text{<msup><msup><mo>y</mo><mo>′</mo></msup><mo>′</mo></msup>=}\frac{{\text{2x}}^3{\text{+2x}}^2{\text{+4x-2x}}^3{\text{-x}}^2}{{{\text{(x}}^2\text{+x+2)}}^2}$

  $\text{<msup><msup><mo>y</mo><mo>′</mo></msup><mo>′</mo></msup>=}\frac{{\text{x}}^2\text{+4x}}{{{\text{(x}}^2\text{+x+2)}}^2}$

  $\text{<msup><msup><mo>y</mo><mo>′</mo></msup><mo>′</mo></msup>=}\frac{\text{x(x+4)}}{{{\text{(x}}^2\text{+x+2)}}^2}$

The reason the solution is useful in providing an answer to the question is that we may now apply the second derivative test for concavity: Namely, a (second differentiable) function f(x)  is concave up (respectively, concave down) at a point c if and only if f''(c)>0

 (respectively, f"(c)<0). Now, let's look at the equation for the second derivative you have above:

  $\text{<msup><msup><mo>y</mo><mo>′</mo></msup><mo>′</mo></msup>=}\frac{\text{x(x+4)}}{{{\text{(x}}^2\text{+x+2)}}^2}$

Note that the denominator of this function is always positive, as a squared real number is either zero or positive, and we can easily use the quadratic formula to see that it is nonzero for xx real. So, to determine whether the second derivative is positive or negative, we need only look at whether the numerator is positive or negative. $\text{x(x+4)=0}$ implies $\text{x=0 or x=-4}$. Plugging values in between these two into $\text{x(x+4)}$ yields that $\text{x(x+4)<0}$  if and only if $\text{-4<x<0}$ , and $\text{x(x+4)>0 }$   when $\text{x<-4 }$ or when $\text{x>0}$ . Hence, applying the second derivative test:

y is concave down when $\text{4<x<0}$ .