Find the point where the tangent line is horizontal in the following function: f(x)=(x−2)(x2−x−11)

Question

Find the point where the tangent line is horizontal in the following function:  f(x)=(x−2)(x2−x−11)

Buck Henry · Accepted Answer

Simplify f′(x)=(x−2)(2x−1)+(1)(x2−x−11)f′(x)=(2x2−5x+2)+(x2−x−11)f′(x)=3x2−6x−9And find where f′(x)=03x2−6x−9=0≤ft⇒x2−2x−3=(x+1)(x−3)=0That&#039;s where slope is 0, hence any line tangent at that point will be horizontal: when x=3 or when x=-1So the roots (x values) of the points you need arex1=3, andx2=−1Then find the corresponding y value in the ORIGINAL equation:f(X)=(x−2)(x2−x−11)to determine the two points: (x1,y1),(x2,y2) where the line tangent to f(x) is horizontal.y1=f(3)=1⋅−5=−5y2=f(−1)=(−3)(−9)=27So your points are (3,−5) and (−1,27)

stomachdm · Accepted Answer

You are on your way there. Consider your first derivativef′(x)=(x−2)(2x−1)+(1)(x2−x−11)Let&#039;s simplify this:f′(x)=(2x2−5x+2)+(x2−x−11)f′(x)=3x2−6x−9Now, for a horizontal line, f′(x)=0. So let&#039;s solve3x2−6x−9=0x2−2x−3=0(x−3)(x+1)=0x=3 or x=−1Hence, you answer that the tangent is horizontal at 2 points, (3,−5) and (−1,27)

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