Question

find f(x) and find the value(s) of x where the tangent line is horizontal. f(x)=x^3(x - 8)^5 f(x)=□

Explanation:

Step1: Apply product - rule

The product - rule states that if \(y = u\cdot v\), then \(y'=u'v + uv'\). Let \(u = x^{3}\) and \(v=(x - 8)^{5}\). First, find \(u'\) and \(v'\). The derivative of \(u=x^{3}\) using the power - rule \((x^{n})'=nx^{n - 1}\) is \(u' = 3x^{2}\). The derivative of \(v=(x - 8)^{5}\) using the chain - rule \((f(g(x)))'=f'(g(x))\cdot g'(x)\) where \(f(u)=u^{5}\), \(u = x - 8\), \(f'(u)=5u^{4}\) and \(g'(x)=1\), so \(v'=5(x - 8)^{4}\).

Step2: Calculate \(f'(x)\)

By the product - rule \(f'(x)=u'v+uv'\), we have \(f'(x)=3x^{2}(x - 8)^{5}+x^{3}\cdot5(x - 8)^{4}\).
Factor out the common factors \(x^{2}(x - 8)^{4}\):

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Step3: Find the values of \(x\) for horizontal tangent

A horizontal tangent line occurs when \(f'(x)=0\). Set \(f'(x)=8x^{2}(x - 8)^{4}(x - 3)=0\).
Using the zero - product property, if \(ab = 0\), then \(a = 0\) or \(b = 0\).
\(x^{2}=0\) gives \(x = 0\) (with multiplicity 2), \((x - 8)^{4}=0\) gives \(x = 8\) (with multiplicity 4), and \(x - 3=0\) gives \(x = 3\).

Answer:

\(f'(x)=8x^{2}(x - 8)^{4}(x - 3)\); The values of \(x\) where the tangent line is horizontal are \(x = 0\), \(x = 3\), and \(x = 8\).