SOLUTION: Determine the intervals of concavity and state any points of inflection.
{{{ f(x) = (1/6)x^4 - (7/6)x^3 + (3/2)x^2 - 8x + 10 }}}.
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Question 1163545: Determine the intervals of concavity and state any points of inflection.
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Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
 = \frac{1}{6}x^4 - \frac{7}{6}x^3+\frac{3}{2}x^2-8x+10)
 = \frac{1}{6}*4x^3 - \frac{7}{6}*3x^2+\frac{3}{2}*2x-8)
First derivative
 = \frac{4}{6}x^3 - \frac{21}{6}x^2+\frac{6}{2}x-8)
 = \frac{2}{3}x^3 - \frac{7}{2}x^2+3x-8)
 = \frac{2}{3}*3x^2 - \frac{7}{2}*2x+3)
Second derivative
 = 2x^2 - 7x+3)
----------------------------
Solve  = 0)
for x to find the two solutions x = 1/2 and x = 3
Plug in a value to the left of x = 1/2, say x = 0, to find that )
is a positive value (the result is positive 3 when you use x = 0), so )
is positive on the interval )
meaning f(x) is concave up on this interval.
Plug in some number between x = 1/2 and x = 3. I'll use x = 1. If you plug x = 1 into the second derivative function, you should get a negative result. This means is negative throughout the interval )
and f(x) is concave down on this interval.
Lastly, f(x) is concave up on the interval )
because plugging in a value to the right of x = 3, say x = 4, leads to being positive.
Because there is a sign change of as we pass through the roots x = 1/2 and x = 3, this means we have points of inflection at these x values. Plug those x values one at a time into f(x) to find the corresponding y coordinates are y = 6.23958 (approx) and y = -18.5
The points of infection are (1/2, 6.23958) and (3, -18.5)
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