calculus

Consider the following.
y = (2x2 + 5)(x3 − 25x)
at (5, 0)
(a) At the indicated point, find the slope of the tangent line.


(b) Find the instantaneous rate of change of the function.

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  1. First check if (5,0) is on the line:
    f(x)= (2x2 + 5)(x3 − 25x)
    f(5)= (50+5)(125-125)=0 indeed.

    a)Slope
    Slope of the tangent line at (5,0) is the value of f'(x) at (5,0), namely f'(5).
    So let's find f'(x).
    We can find f'(x) as a product, or as an expanded polynomial. I choose the latter:
    y = (2x2 + 5)(x3 − 25x)
    =2x^5+5x^3-50x^3-125x^2
    =2x^5-45x^3-125x^2
    Differentiate term by term:
    f'(x)=10x^4-135x2-250x
    so
    f'(5)=6250-3375-1250=1625
    = slope of tangent at (5,0)

    b.
    instantaneous rate of change of the function is precisely f'(x)=dy/dx.
    So the answer is the same as in a.

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