using the definition of the slope of a tangent line, and find the slope of each function at x=c

Question

f(x)=3-7x at x=-4

g(x)=2x^2+x. at x=0

h(x)= 1/3x. at x=6

Answer 1

The limit definition of slope is:

So for f(x) at x = -4:

f(x + h) = f(-4 + h) = 3 - 7(-4 + h) = 3 + 28 - 7h = 31 - 7h

f(x) = f(-4) = 3 - 7(-4) = 31

f(x + h) - f(x) = 31 - 7h - (31) = -7h

(f(x + h) - f(x))/h = -7h/h = -7 so the limit as h→0 is also -7

For g(x) at x = 0:

g(x + h) = 2(x + h)² + (x + h) = 2(0 + h)² + (0 + h) = 2h² + h

g(x) = g(0) = 2(0)² + (0) = 0

g(x + h) - g(x) = 2h² + h - 0 = 2h² + h

(g(x + h) - g(x))/h = (2h² + h)/h = 2h + 1 so the limit as h→0 is 1

I'll let you do the last problem since I don't know if it is h(x) = (1/3)•x or 1/(3x) but you just do the same thing. The answer will be -1/108