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vredina [299]
4 years ago
11

Find all zeros of function and write the polynomial as a product of linear factorsX^3-x^2+9x-9

Mathematics
1 answer:
mart [117]4 years ago
8 0

Answer:

\large\boxed{x=1\ \vee\ x=-3i\ \vee\ x=3i}

Step-by-step explanation:

\text{The zeros:}\\\\x^3-x^2+9x-9=0\\\\x^2(x-1)+9(x-1)=0\\\\(x-1)(x^2+9)=0\iff x-1=0\ \vee\ x^2+9=0\\\\x-1=0\qquad\text{add 1 to both sides}\\x-1+1=0+1\\\boxed{x=1}\\\\x^2+9=0\qquad\text{subtract 9 from both sides}\\x^2=-9

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(a) mtan refers to the slope of the tangent line. Given <em>f(x)</em> = 9 + 7<em>x</em> ², compute the difference quotient:

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Then as <em>h</em> approaches 0 - bearing in mind that we're specifically considering <em>h</em> <em>near</em> 0, and not <em>h</em> = 0 - we can eliminate the factor of <em>h</em> in the numerator and denominator, so that

m_{\rm tan} = \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \lim_{h\to0}\frac{14xh+7h^2}h = \lim_{h\to0}(14x+7h) = 14x

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