Question Watch Video Show Examples Given the function $f(x)=x^{4}-8 x^{3}+18 x^{2}$, determine all intervals on which $f$ is both decreasing and concave up. Answer Attempt 2 out of 2

Step 1 ：Find the first derivative of the function $f(x)$: $f'(x) = 4x^{3}-24x^{2}+36x$

Step 2 ：Set $f'(x)$ less than 0 and solve for $x$ to find where the function is decreasing: $4x^{3}-24x^{2}+36x < 0$

Step 3 ：Simplify the inequality by dividing through by 4: $x^{3}-6x^{2}+9x < 0$

Step 4 ：Factor out an $x$: $x(x^{2}-6x+9) < 0$

Step 5 ：Further factor the expression: $x(x-3)^{2} < 0$

Step 6 ：Set each factor equal to 0 to find $x = 0$ and $x = 3$

Step 7 ：Find the second derivative of the function $f(x)$: $f''(x) = 12x^{2}-48x+36$

Step 8 ：Set $f''(x)$ greater than 0 and solve for $x$ to find where the function is concave up: $12x^{2}-48x+36 > 0$

Step 9 ：Simplify the inequality by dividing through by 12: $x^{2}-4x+3 > 0$

Step 10 ：Factor the expression: $(x-1)(x-3) > 0$

Step 11 ：Set each factor equal to 0 to find $x = 1$ and $x = 3$

Step 12 ：Find the intersection of the intervals where the function is decreasing and where it is concave up. The function is decreasing for $x < 0$ and $0 < x < 3$. The function is concave up for $x < 1$ and $x > 3$. The only interval where the function is both decreasing and concave up is $0 < x < 1$

Step 13 ：\(\boxed{\text{The function } f(x)=x^{4}-8 x^{3}+18 x^{2} \text{ is decreasing and concave up on the interval } (0, 1)}\)