consider the function\n\n$f(x)=x^{3}-4.5x^{2}-12x - 1$.\n\na) determine the intervals on which $f$ is…

consider the function\n\n$f(x)=x^{3}-4.5x^{2}-12x - 1$.\n\na) determine the intervals on which $f$ is concave up and concave down.\n$f$ is concave up on: \n$f$ is concave down on: \n\nb) based on your answer to part (a), determine the inflection points of $f$. each point should be entered as an ordered pair (that\nis, in the form $(x,y)$).\n(separate multiple answers by commas.)\n\nnote: when using interval notation in webwork, remember that:\n you use inf for $\\infty$ and -inf for $-\\infty$,\n and use u for the union symbol.\nenter dne if an answer does not exist.

consider the function\n\n$f(x)=x^{3}-4.5x^{2}-12x - 1$.\n\na) determine the intervals on which $f$ is concave up and concave down.\n$f$ is concave up on: \n$f$ is concave down on: \n\nb) based on your answer to part (a), determine the inflection points of $f$. each point should be entered as an ordered pair (that\nis, in the form $(x,y)$).\n(separate multiple answers by commas.)\n\nnote: when using interval notation in webwork, remember that:\n you use inf for $\\infty$ and -inf for $-\\infty$,\n and use u for the union symbol.\nenter dne if an answer does not exist.

Answer

Explanation:

Step 1: Find the first - derivative

Using the power rule ((x^n)^\prime=nx^{n - 1}), for (f(x)=x^{3}-4.5x^{2}-12x - 1), we have (f^\prime(x)=3x^{2}-9x - 12).

Step 2: Find the second - derivative

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

Step 3: Find the critical points of (f^{\prime\prime}(x))

Set (f^{\prime\prime}(x)=0), then (6x - 9=0). Solving for (x), we get (x=\frac{9}{6}=\frac{3}{2}).

Step 4: Test the intervals for concavity

  • Interval ((-\infty,\frac{3}{2})): Let (x = 0), then (f^{\prime\prime}(0)=6\times0 - 9=-9<0). So (f(x)) is concave down on ((-\infty,\frac{3}{2})).
  • Interval ((\frac{3}{2},\infty)): Let (x = 2), then (f^{\prime\prime}(2)=6\times2 - 9 = 3>0). So (f(x)) is concave up on ((\frac{3}{2},\infty)).

Step 5: Find the inflection point

Substitute (x = \frac{3}{2}) into (f(x)). (f(\frac{3}{2})=(\frac{3}{2})^{3}-4.5\times(\frac{3}{2})^{2}-12\times\frac{3}{2}-1) (=\frac{27}{8}-4.5\times\frac{9}{4}-18 - 1) (=\frac{27}{8}-\frac{81}{8}-19) (=\frac{27 - 81}{8}-19) (=-\frac{54}{8}-19) (=-\frac{27}{4}-19=-\frac{27 + 76}{4}=-\frac{103}{4}=-25.75).

Answer:

a) (f) is concave up on: ((\frac{3}{2},\text{INF})) (f) is concave down on: ((-\text{INF},\frac{3}{2})) b) ((\frac{3}{2},-\frac{103}{4}))