Question

find the critical numbers of the function ( f(x)=12 x^{5}-45 x^{4}-20 x^{3}-4 ) and classify them using a graph.
( x=0 ) is a select an answer
( x= ) is a select an answer
enter an integer or decimal number more...
( x= ) is a select an answer
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Explanation:

Step1: Find the derivative of the function

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(y = f(x)=12x^{5}-45x^{4}-20x^{3}-4\), the derivative \(f^\prime(x)=60x^{4}-180x^{3}-60x^{2}=60x^{2}(x^{2}-3x - 1)\)

Step2: Find the critical numbers

Set \(f^\prime(x)=0\).

• Case 1: \(60x^{2}=0\), then \(x = 0\)
• Case 2: For \(x^{2}-3x - 1=0\), using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 1\), \(b=-3\), \(c=-1\).

\(x=\frac{3\pm\sqrt{(-3)^{2}-4\times1\times(-1)}}{2\times1}=\frac{3\pm\sqrt{9 + 4}}{2}=\frac{3\pm\sqrt{13}}{2}\approx\frac{3\pm3.606}{2}\)
\(x_1=\frac{3 + 3.606}{2}\approx3.303\), \(x_2=\frac{3-3.606}{2}\approx - 0.303\)

Step3: Analyze the sign of the derivative around the critical numbers (using a graph - conceptually)

• For \(x\lt - 0.303\), pick \(x=-1\), \(f^\prime(-1)=60\times(-1)^{2}\times((-1)^{2}-3\times(-1)-1)=60\times(1 + 3-1)=180>0\)
• For \(-0.303\lt x\lt0\), pick \(x=-0.1\), \(f^\prime(-0.1)=60\times(-0.1)^{2}\times((-0.1)^{2}-3\times(-0.1)-1)=60\times0.01\times(0.01 + 0.3-1)\lt0\)
• For \(0\lt x\lt3.303\), pick \(x = 1\), \(f^\prime(1)=60\times1^{2}\times(1^{2}-3\times1-1)=60\times(1-3 - 1)\lt0\)
• For \(x>3.303\), pick \(x = 4\), \(f^\prime(4)=60\times4^{2}\times(4^{2}-3\times4-1)=60\times16\times(16-12 - 1)>0\)

Since \(f^\prime(x)\) changes sign from positive to negative at \(x\approx - 0.303\), \(x\approx - 0.303\) is a local maximum.
Since \(f^\prime(x)\) does not change sign at \(x = 0\) (the sign of \(f^\prime(x)\) is negative on both sides of \(x = 0\) in the non - zero neighborhood), \(x = 0\) is not a local maximum or minimum (a point of inflection in terms of derivative behavior for critical point classification here).
Since \(f^\prime(x)\) changes sign from negative to positive at \(x\approx3.303\), \(x\approx3.303\) is a local minimum

Answer:

\(x = 0\) is a neither local maximum nor local minimum (or a point where the derivative has a zero but no sign change for local extrema).
\(x\approx - 0.30\) is a local maximum.
\(x\approx3.30\) is a local minimum.