hw 18 - second derivative test section 3.4: problem 3\n(1 point)\nconsider the function\n$f(x)=x^{3}-4.5x^{2}…

hw 18 - second derivative test section 3.4: problem 3\n(1 point)\nconsider the function\n$f(x)=x^{3}-4.5x^{2}-12x + 3$.\n a) 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 b) based on your answer to part (a), determine the inflection points of $f$. each point should be entered as an ordered pair (that\n is, in the form $(x,y)$).\n(separate multiple answers by commas.)\n c) find the critical numbers of $f$ and use the second derivative test, when possible, to determine the relative extrema. list only\nthe $x$-coordinates.\n relative maxima at: (separate multiple answers by commas.)\n relative minima at: (separate multiple answers by commas.)\n(round to three decimal places at needed.)

hw 18 - second derivative test section 3.4: problem 3\n(1 point)\nconsider the function\n$f(x)=x^{3}-4.5x^{2}-12x + 3$.\n a) 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 b) based on your answer to part (a), determine the inflection points of $f$. each point should be entered as an ordered pair (that\n is, in the form $(x,y)$).\n(separate multiple answers by commas.)\n c) find the critical numbers of $f$ and use the second derivative test, when possible, to determine the relative extrema. list only\nthe $x$-coordinates.\n relative maxima at: (separate multiple answers by commas.)\n relative minima at: (separate multiple answers by commas.)\n(round to three decimal places at needed.)

Answer

Explanation:

Step1: Find the first and second derivatives

The first derivative (f^\prime(x)=3x^{2}-9x - 12) (using the power rule ((x^n)^\prime=nx^{n - 1})). The second derivative (f^{\prime\prime}(x)=6x-9).

Step2: Find the inflection point

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

Step3: Determine concavity

  • For (x>\frac{3}{2}), let (x = 2). Then (f^{\prime\prime}(2)=6\times2 - 9=3>0). So (f(x)) is concave up on ((\frac{3}{2},\infty)).
  • For (x<\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})).

Step4: Find the inflection - point (y) - value

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}+3) (=\frac{27}{8}-4.5\times\frac{9}{4}-18 + 3) (=\frac{27}{8}-\frac{81}{8}-15) (=\frac{27 - 81}{8}-15) (=-\frac{54}{8}-15=-\frac{27}{4}-15=-\frac{27 + 60}{4}=-\frac{87}{4}=-21.75). So the inflection point is ((\frac{3}{2},-21.75)).

Step5: Find critical numbers

Set (f^\prime(x)=0), (3x^{2}-9x - 12 = 0). Divide by (3): (x^{2}-3x - 4=0). Factor: ((x + 1)(x - 4)=0). So (x=-1) and (x = 4).

Step6: Use the second - derivative test

  • For (x=-1), (f^{\prime\prime}(-1)=6\times(-1)-9=-15<0). So (x=-1) is a relative maximum.
  • For (x = 4), (f^{\prime\prime}(4)=6\times4-9=15>0). So (x = 4) is a relative minimum.

Answer:

a) (f) is concave up on: ((\frac{3}{2},\infty)); (f) is concave down on: ((-\infty,\frac{3}{2})) b) ((\frac{3}{2},-21.75)) c) Relative maxima at: (-1); Relative minima at: (4)