Question 1163545
*[Tex \Large f(x) = \frac{1}{6}x^4 - \frac{7}{6}x^3+\frac{3}{2}x^2-8x+10]


*[Tex \Large f'(x) = \frac{1}{6}*4x^3 - \frac{7}{6}*3x^2+\frac{3}{2}*2x-8] First derivative


*[Tex \Large f'(x) = \frac{4}{6}x^3 - \frac{21}{6}x^2+\frac{6}{2}x-8]


*[Tex \Large f'(x) = \frac{2}{3}x^3 - \frac{7}{2}x^2+3x-8]


*[Tex \Large f^{\prime \prime}(x) = \frac{2}{3}*3x^2 - \frac{7}{2}*2x+3] Second derivative


*[Tex \Large f^{\prime \prime}(x) = 2x^2 - 7x+3] 


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Solve *[Tex \Large f^{\prime \prime}(x) = 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 *[Tex \Large f^{\prime \prime}(0)] is a positive value (the result is positive 3 when you use x = 0), so *[Tex \Large f^{\prime \prime}(x)] is positive on the interval *[Tex \Large (-\infty, 1/2)] 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 *[Tex \Large f^{\prime \prime}(x)] is negative throughout the interval *[Tex \Large (1/2, 3)] and f(x) is concave down on this interval.


Lastly, f(x) is concave up on the interval *[Tex \Large (3, \infty)] because plugging in a value to the right of x = 3, say x = 4, leads to *[Tex \Large f^{\prime \prime}(x)] being positive.


Because there is a sign change of *[Tex \Large f^{\prime \prime}(x)] 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)