Surnmarize the pertinent information obtained by \( f(x)=2 x(x-3)^{3} \) On what interval(s) is \( f \) concave upward or conca (Type your answer in interval notation. Use integ A. \( f \) is concave upward on \( \square \) and conceve B. \( f \) is never concave downward, \( f \) is concav C. \( f \) is never concave upward; \( f \) is concave \( d \)

1. Write the function as \[ f(x)=2x(x-3)^3. \] 2. Compute the first derivative using the product rule (with \( u=2x \) and \( v=(x-3)^3 \)): \[ f'(x)=2(x-3)^3+2x\cdot3(x-3)^2=2(x-3)^2\bigl[(x-3)+3x\bigr]. \] Simplify the bracket: \[ (x-3)+3x=4x-3. \] Thus, \[ f'(x)=2(x-3)^2(4x-3). \] 3. Compute the second derivative. Write: \[ f'(x)=2(x-3)^2(4x-3). \] Use the product rule letting \( u=2(x-3)^2 \) and \( v=4x-3 \). Then, \[ u'=4(x-3) \quad \text{and} \quad v'=4. \] So, \[ f''(x)=u'v+uv'=4(x-3)(4x-3)+2(x-3)^2\cdot4. \] Simplify: \[ f''(x)=4(x-3)(4x-3)+8(x-3)^2. \] Factor out \( 4(x-3) \): \[ f''(x)=4(x-3)\Bigl[(4x-3)+2(x-3)\Bigr]. \] Expand the expression inside the bracket: \[ (4x-3)+2(x-3)=4x-3+2x-6=6x-9. \] Factor \( 6x-9 \): \[ 6x-9=3(2x-3). \] Therefore, \[ f''(x)=12(x-3)(2x-3). \] 4. Find possible inflection points by setting \( f''(x)=0 \): \[ 12(x-3)(2x-3)=0. \] Solve: - \( x-3=0 \) gives \( x=3 \), - \( 2x-3=0 \) gives \( x=\frac{3}{2} \). 5. Determine the intervals of concavity by testing the sign of \( f''(x) \) in the intervals determined by \( x=\frac{3}{2} \) and \( x=3 \): - For \( x<\frac{3}{2} \), choose \( x=0 \): \[ f''(0)=12(0-3)(2\cdot0-3)=12(-3)(-3)=12\cdot9>0. \] So, \( f \) is concave upward on \((-\infty,\frac{3}{2})\). - For \( \frac{3}{2}3 \), choose \( x=4 \): \[ f''(4)=12(4-3)(2\cdot4-3)=12(1)(5)=60>0. \] So, \( f \) is concave upward on \((3,\infty)\). 6. Summarize the results in interval notation: - \( f \) is concave upward on \((-\infty,\frac{3}{2})\cup(3,\infty) \). - \( f \) is concave downward on \((\frac{3}{2},3) \).