Question

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

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: \(4\)