determine the open intervals on which the graph of the function is concave upward or concave downward…

determine the open intervals on which the graph of the function is concave upward or concave downward. (enter your answers using interval notation. if an answer does not exist, enter dne.)\n\nf(x) = x⁴ - 9x³\n\nconcave upward\n\nconcave downward

Answer

Explanation:

Step1: Find the first derivative

Using the power rule ((x^n)^\prime=nx^{n - 1}), for (y = f(x)=x^{4}-9x^{3}), we have (f^\prime(x)=4x^{3}-27x^{2}).

Step2: Find the second derivative

Differentiate (f^\prime(x)) again. (f^{\prime\prime}(x)=(4x^{3}-27x^{2})^\prime = 12x^{2}-54x=6x(2x - 9)).

Step3: Find the critical points of (f^{\prime\prime}(x))

Set (f^{\prime\prime}(x)=0), so (6x(2x - 9)=0). Solving (6x(2x - 9)=0) gives (x = 0) and (x=\frac{9}{2}).

Step4: Test the intervals

For the interval ((-\infty,0)), let (x=-1). Then (f^{\prime\prime}(-1)=12\times(-1)^{2}-54\times(-1)=12 + 54=66>0).
For the interval ((0,\frac{9}{2})), let (x = 1). Then (f^{\prime\prime}(1)=12\times1^{2}-54\times1=12-54=-42<0).
For the interval ((\frac{9}{2},\infty)), let (x = 5). Then (f^{\prime\prime}(5)=12\times5^{2}-54\times5=300 - 270 = 30>0).

Answer:

concave upward: ((-\infty,0)\cup(\frac{9}{2},\infty)) concave downward: ((0,\frac{9}{2}))