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.

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.

To find the slope of the tangent line at the indicated point (5, 0) for the given function y = (2x^2 + 5)(x^3 − 25x), we need to find the derivative of the function and then substitute x = 5 into the derivative.

(a) To find the derivative of the function, we can use the product rule of differentiation. The product rule states that for two functions u(x) and v(x), the derivative of their product is given by:

d/dx(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)

Applying the product rule to our function, we have:

y = (2x^2 + 5)(x^3 − 25x)
Taking the derivative with respect to x, we get:

dy/dx = (2x^2 + 5)(3x^2) + (2x^2 + 5)(-25)

Simplifying this expression, we have:

dy/dx = 6x^4 + 15x^2 - 50x^2 - 125

Simplifying further, we get:

dy/dx = 6x^4 - 35x^2 - 125

Now we need to substitute x = 5 into this derivative expression to find the slope of the tangent line at the point (5, 0):

Slope at (5, 0) = dy/dx evaluated at x = 5

Substituting x = 5, we get:

Slope at (5, 0) = 6(5^4) - 35(5^2) - 125

Evaluating this expression, we find the slope of the tangent line at the point (5, 0).

(b) To find the instantaneous rate of change of the function, we need to evaluate the derivative at the given point (5, 0), which we have already done in part (a).

So, the instantaneous rate of change of the function at the point (5, 0) is given by the value of the derivative at x = 5.