consider the following. (if an answer does not exist, enter dne.) f(x) = 2x^3 - 9x^2 + 12x - 4 find the…

Explanation:

Step1: Find the first - derivative

Differentiate $f(x)=2x^{3}-9x^{2}+12x - 4$ using the power rule. $f'(x)=6x^{2}-18x + 12$

Step2: Find the second - derivative

Differentiate $f'(x)$ using the power rule. $f''(x)=12x-18$

Step3: Find the inflection point

Set $f''(x) = 0$ and solve for $x$. $12x-18 = 0$ $12x=18$ $x=\frac{18}{12}=\frac{3}{2}$ To find the $y$ - coordinate of the inflection point, substitute $x = \frac{3}{2}$ into $f(x)$. $f(\frac{3}{2})=2(\frac{3}{2})^{3}-9(\frac{3}{2})^{2}+12(\frac{3}{2})-4$ $=2\times\frac{27}{8}-9\times\frac{9}{4}+18 - 4$ $=\frac{27}{4}-\frac{81}{4}+14$ $=\frac{27 - 81}{4}+14$ $=-\frac{54}{4}+14$ $=-\frac{27}{2}+\frac{28}{2}=\frac{1}{2}$ So the inflection point is $(\frac{3}{2},\frac{1}{2})$

Step4: Determine concavity intervals

Test intervals based on the inflection - point $x=\frac{3}{2}$. Choose a test point in the interval $(-\infty,\frac{3}{2})$, say $x = 1$. Then $f''(1)=12\times1-18=-6<0$, so $f(x)$ is concave down on $(-\infty,\frac{3}{2})$. Choose a test point in the interval $(\frac{3}{2},\infty)$, say $x = 2$. Then $f''(2)=12\times2-18 = 6>0$, so $f(x)$ is concave up on $(\frac{3}{2},\infty)$

Answer:

Concave up: $(\frac{3}{2},\infty)$ Concave down: $(-\infty,\frac{3}{2})$ Inflection point: $(\frac{3}{2},\frac{1}{2})$