Posted by **Adam Janky ** on Thursday, March 10, 2011 at 9:53pm.

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.

- calculus -
**MathMate**, Friday, March 11, 2011 at 8:07am
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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