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20 June, 13:37

Find the slope and equation of the tangent line to the graph of the function at the given value of x.

f (x) = x^4-20x^2+64; x=-1

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  1. 20 June, 13:40
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    The slope is the differential of the function.

    Recall, if y = x^n, (dy/dx) = nx^ (n-1).

    y = x^4-20x^2+64; x = - 1. To differentiate this, we do it for each term.

    (dy/dx) = (4) (x^ (4 - 1)) - (2) (20x^ (2-1) + 0*64x^ (0-1) (Note 64 = 64x^0, x^0 = 1)

    = (4) x^ (3) - 40x^ (1) + 0

    = 4x^3 - 40x^1.

    (dy/dx) = 4x^3 - 40x. Note at x = - 1.

    (dy/dx), x = - 1, = 4 (-1) ^3 - 40 (-1)

    = - 4 + 40 = 40 - 4 = 36.

    Slope at x = - 1 is 36.

    Cheers.
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