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Horizontal Tangent At what point is the tangent to \(f(x)=3-4 x-x^{2}\) horizontal?

Short Answer

Expert verified
The point at which the tangent to the function \(f(x)=3-4x-x^2\) is horizontal is (2, -5).

Step by step solution

01

Find the Derivative

The derivative of \(f(x)=3-4x-x^2\) is given by \(f'(x)=-4-2x\).
02

Determine where the derivative is zero

Setting the derivative equal to zero gives us the equation \(-4-2x = 0\). Solving this equation for \(x\) gives \(x=-4/-2=2\).
03

Find the \(y\)-coordinate

Substitute \(x=2\) into the original function to get the \(y\)-coordinate: \(f(2)=3-4*2-2^2=-5\).
04

Write down the point

The point at which the tangent line to \(f(x)=3-4x-x^2\) is horizontal is thus (2, -5).

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