Calculus 1 Question Derivatives

Question

1)Assume that g(y) has a derivative at y = 3.5 with g′(−3.5) = −3. Using the definition of the derivative, explain why the function f (y) = g(y) − 10 has a derivative at y = 3.5 with

f′(3.5)=−3.

2)Assume that g(y) has a derivative at y = 3.5 with g′(−3.5) = −3. Use geometry to explain why the function f (y) = g(y) − 10 has a derivative at y = 3.5 with f ′(3.5) = −3.

Answer 1

The - part of the -3.5 is a typo. For 1, use the limit definition of derivative at a pt: f'(a) = lim_x→a (f(x)-f(a))/(x-a) :

f(y) = g(y) - 10. , f(3.5) = g(3.5) - 10

g'(3.5) = lim_y→3.5 (g(y) - g(3.5)) / (y - 3.5) = - 3

f'(3.5) = lim_y→3.5 (f(y) - f(3.5)) / (y - 3.5) = (g(y) - 10 - (g(3.5) - 10))/ (y - 3.5) = (g(y) - g(3.5)) / (y - 3.5) = - 3 ◊

2) The graph of f(y) is the graph of g(y) shifted down 10, which will not alter its shape (rate of change). The tangent to f's curve at y = 3.5 will have the same slope as the tangent to g's curve at y = 3.5.