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Given the function $f(x)=x^4-8x^3+18x^2$, 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 $f(x)$: $f^{\prime }(x)=4x^3-24x^2+36x$

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