The tangent line to a function f(x) at x=4 is found to be y= -5x+6.
Since the point of tangency is on both f(x) and the tangent line,
and since when x=4, the tangent line and x=4 has y value
y = -5(4)+6 = -20+6 = -14,
then (4,-14) the point on tangency is a point on the
graph of f(x). That is,
f(4) = -14
Since f'(4) is the slope of the tangent line at x=4, and the tangent
line is y = -5x+6 which has slope -5 (which is found either by
comparing it to y = mx+b, or by finding
).
then f'(4) = 5
g(x) = 2f(x)+8
so
g(4) = 2f(4)+8
g(4) = 2(-14)+8
g(4) = -28+8
g(4) = -20
Since
g(x) = 2f(x)+8,
taking the derivative of both sides:
g'(x) = 2f'(x)
g'(4) = 2f'(4)
g'(4) = 2(5)
g'(4) = 10
Edwin