Question

consider the function

$f(x)=x^{3}-4.5x^{2}-12x - 1$.

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.)

note: when using interval notation in webwork, remember that:
you use inf for $\infty$ and -inf for $-\infty$,
and use u for the union symbol.
enter dne if an answer does not exist.

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) \((\frac{3}{2},-\frac{103}{4})\)