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NUMBER 7Inc. intervals = increasing intervalDec. intervals = decreasing intervals

NUMBER 7Inc. intervals = increasing intervalDec. intervals = decreasing intervals-example-1

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\begin{gathered} \text{Domain: (-}\infty,\infty) \\ \text{Range: \lbrack-5.848, }\infty) \\ relative\text{ minimum at x = -0.366} \\ \text{relative max i}mum\text{ at x =1}.366 \\ \text{Increasing interval: (-0.366},\text{ 1.366) }\cup(2,\text{ }\infty) \\ \text{Decreasing interval: (-}\infty,\text{ -0.366)} \end{gathered}

See explanation below

Step-by-step explanation:


\begin{gathered} 7)\text{ The domain: x values of the function} \\ \text{The domain of the polynomial is all real numbers} \\ \text{Domain: (-}\infty,\infty) \end{gathered}
\begin{gathered} \text{Range: y values of the function} \\ \text{The range starts before -6 and extends towards infinity} \\ On\text{ the graph, this point is y = -5.}848 \\ \text{Range is y }\ge\text{ -5.}848 \\ \text{Range: \lbrack-5.848, }\infty) \end{gathered}
\begin{gathered} To\text{ }get\operatorname{Re}l\text{ a tive maximum and minimum, we differenciate the initial polynomial:} \\ f(x)=x^4-4x^3+3x^2+4x-5 \\ f^(\prime)(x)=4x^3-12x^2\text{ + 6x + 4} \\ 4x^3-12x^2\text{ + 6x + 4 = 0} \\ \text{The roots of the }4x^3-12x^2\text{ + 6x + 4 are }-0.366,\text{ 1.366, }2 \end{gathered}
\begin{gathered} \text{second derivative: }f^(\prime\prime)(x)=12x^2\text{ - 24x + 6} \\ \text{when x = }-0.366 \\ \text{roots of }12x^2\text{ - 24x + 6 = 16.39} \\ \text{when x = 1.366} \\ \text{roots of }12x^2\text{ - 24x + 6 =}-4.39 \\ \text{when x = 2} \\ \text{roots of }12x^2\text{ - 24x + 6 = 6} \end{gathered}
\begin{gathered} if\text{ after inserting the the values above:} \\ \text{the 2nd derivative result is positive, it will be local minimum} \\ \text{the 2nd derivative result is negative, it will be local max imum} \\ relative\text{ minimum at x = -0.366} \\ \text{relative max i}mum\text{ at x =1}.366 \end{gathered}
\begin{gathered} \text{End behaviour: }f(x)=x^4-4x^3+3x^2+4x-5 \\ \text{leading coefficient = 1 (positive), degr}ee\text{ = 4 (even)} \\ As\text{ x }\rightarrow\text{ +}\infty,\text{ f(x) }\rightarrow\text{ +}\infty \\ As\text{ x }\rightarrow\text{ -}\infty,\text{ f(x) }\rightarrow\text{ +}\infty \end{gathered}
\begin{gathered} \text{Increasing interval:} \\ =\text{ (-0.366},\text{ 1.366) }\cup(2,\text{ }\infty) \\ \text{Decreasing inrterval}\colon \\ =\text{ (-}\infty,\text{ -0.366)} \end{gathered}

NUMBER 7Inc. intervals = increasing intervalDec. intervals = decreasing intervals-example-1
User Sander De Dycker
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