# Find an equation for the tangent line to the graph of f(x)= (2/3)-2x-x^2-(1/3)x^3 having maximum slope. The slope of the tangent line to a curve at a point is the value of the derivative of the function at that point (if the derivative exists.)

Given `f(x)=2/3-2x-x^2-1/3x^3` :

`f'(x)=-2-2x-x^2`

The derivative function is quadratic; its graph is a parabola which opens down so it has an absolute...

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The slope of the tangent line to a curve at a point is the value of the derivative of the function at that point (if the derivative exists.)

Given `f(x)=2/3-2x-x^2-1/3x^3` :

`f'(x)=-2-2x-x^2`

The derivative function is quadratic; its graph is a parabola which opens down so it has an absolute maximum at the vertex. We can use calculus to find the maximum by taking the derivative of `f'(x)` :

`f''(x)=-2-2x` The maximum will occur when `f''(x)=0==>x=-1`

At x=-1, the slope of the tangent line is `f'(-1)=-1`

The tangent line intersects the curve at `(x,f(x))=(-1,2)`

The equation of the tangent line is `y-2=-1(x+1)` or `y=-x+1`

The graph of the function and its derivative:

It is clear that any other tangent line has a slope that is more negative, thus a slope of -1 is the greatest possible.

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