Question

the function $f(x)=2x^{3}-24x^{2}+72x + 3$ has derivative $f(x)=6x^{2}-48x + 72$.
$f(x)$ has one local minimum and one local maximum.
$f(x)$ has a local minimum at $x$ equals
with value
and a local maximum at $x$ equals
with value

Explanation:

Step1: Find critical points

Set \(f^{\prime}(x)=0\), so \(6x^{2}-48x + 72=0\). Divide through by \(6\): \(x^{2}-8x + 12=0\). Factor: \((x - 2)(x - 6)=0\). Critical points \(x = 2\) and \(x=6\).

Step2: Use second - derivative test

Find \(f^{\prime\prime}(x)\). Differentiate \(f^{\prime}(x)=6x^{2}-48x + 72\), \(f^{\prime\prime}(x)=12x-48\).

• For \(x = 2\): \(f^{\prime\prime}(2)=12\times2-48=-24<0\), so \(x = 2\) is a local maximum.
• For \(x = 6\): \(f^{\prime\prime}(6)=12\times6-48 = 24>0\), so \(x = 6\) is a local minimum.

Step3: Find function values

• For local minimum (\(x = 6\)): \(f(6)=2\times6^{3}-24\times6^{2}+72\times6 + 3=2\times216-24\times36+432 + 3=432-864+432 + 3=3\).
• For local maximum (\(x = 2\)): \(f(2)=2\times2^{3}-24\times2^{2}+72\times2+3=2\times8-24\times4 + 144+3=16-96+144 + 3=67\).

Answer:

Local minimum at \(x = 6\) with value \(3\); local maximum at \(x = 2\) with value \(67\).