To determine where the function is concave upward or concave downward, we need to find the second derivative of the function \( f(x) = \frac{1}{x^2 + 8} \). The function is concave upward where the second derivative is positive and concave downward where it is negative. We will calculate the second derivative, find its critical points, and analyze the sign of the second derivative in the intervals determined by these critical points.
To determine where the function is concave upward or downward, we need to find the second derivative of the function \( f(x) = \frac{1}{x^2 + 8} \).
First, find the first derivative \( f'(x) \) using the quotient rule. The quotient rule states that if \( f(x) = \frac{u(x)}{v(x)} \), then
\[
f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\]
For \( f(x) = \frac{1}{x^2 + 8} \), we have \( u(x) = 1 \) and \( v(x) = x^2 + 8 \).
- \( u'(x) = 0 \)
- \( v'(x) = 2x \)
Applying the quotient rule:
\[
f'(x) = \frac{0 \cdot (x^2 + 8) - 1 \cdot 2x}{(x^2 + 8)^2} = \frac{-2x}{(x^2 + 8)^2}
\]
Next, find the second derivative \( f''(x) \). Use the quotient rule again on \( f'(x) = \frac{-2x}{(x^2 + 8)^2} \).
Let \( u(x) = -2x \) and \( v(x) = (x^2 + 8)^2 \).
- \( u'(x) = -2 \)
- \( v'(x) = 2(x^2 + 8)(2x) = 4x(x^2 + 8) \)
Applying the quotient rule:
\[
f''(x) = \frac{(-2)(x^2 + 8)^2 - (-2x)(4x(x^2 + 8))}{(x^2 + 8)^4}
\]
Simplify the expression:
\[
f''(x) = \frac{-2(x^2 + 8)^2 + 8x^2(x^2 + 8)}{(x^2 + 8)^4}
\]
\[
= \frac{-2(x^2 + 8)^2 + 8x^4 + 64x^2}{(x^2 + 8)^4}
\]
Simplify the numerator:
\[
-2(x^2 + 8)^2 = -2(x^4 + 16x^2 + 64)
\]
\[
= -2x^4 - 32x^2 - 128
\]
Combine terms:
\[
f''(x) = \frac{-2x^4 - 32x^2 - 128 + 8x^4 + 64x^2}{(x^2 + 8)^4}
\]
\[
= \frac{6x^4 + 32x^2 - 128}{(x^2 + 8)^4}
\]
The function is concave upward where \( f''(x) > 0 \) and concave downward where \( f''(x) < 0 \).
Set \( f''(x) = 0 \) to find critical points:
\[
6x^4 + 32x^2 - 128 = 0
\]
Divide by 2:
\[
3x^4 + 16x^2 - 64 = 0
\]
Let \( y = x^2 \), then:
\[
3y^2 + 16y - 64 = 0
\]
Solve using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 3 \), \( b = 16 \), \( c = -64 \).
\[
y = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 3 \cdot (-64)}}{2 \cdot 3}
\]
\[
y = \frac{-16 \pm \sqrt{256 + 768}}{6}
\]
\[
y = \frac{-16 \pm \sqrt{1024}}{6}
\]
\[
y = \frac{-16 \pm 32}{6}
\]
\[
y_1 = \frac{16}{6} = \frac{8}{3}, \quad y_2 = \frac{-48}{6} = -8
\]
Since \( y = x^2 \), \( x^2 = \frac{8}{3} \) gives \( x = \pm \sqrt{\frac{8}{3}} \).
Test intervals around \( x = \pm \sqrt{\frac{8}{3}} \) to determine concavity:
- For \( x < -\sqrt{\frac{8}{3}} \), choose \( x = -3 \).
- For \( -\sqrt{\frac{8}{3}} < x < \sqrt{\frac{8}{3}} \), choose \( x = 0 \).
- For \( x > \sqrt{\frac{8}{3}} \), choose \( x = 3 \).
Calculate \( f''(x) \) for these values to determine the sign.
Concave upward: \( (-\infty, -\sqrt{\frac{8}{3}}) \cup (\sqrt{\frac{8}{3}}, \infty) \)
Concave downward: \( (-\sqrt{\frac{8}{3}}, \sqrt{\frac{8}{3}}) \)
\[
\boxed{\text{Concave upward: } (-\infty, -\sqrt{\frac{8}{3}}) \cup (\sqrt{\frac{8}{3}}, \infty)}
\]
\[
\boxed{\text{Concave downward: } (-\sqrt{\frac{8}{3}}, \sqrt{\frac{8}{3}})}
\]
Answered
