If lim (h->0) ( f(3+h) - f(3) ) / h = 0, then which of the following must be true?

Question

lim   ( f(3+h) - f(3) ) / h = 0

h->0

 

1. I : f has a derivative at x=3
2. II : f is continuous at x=3
3. III : f has a critical value at x=3

I Only

II Only

I and II

I and III

I, II, and III

Can you please explain also?

Answer 1

I, II and III

the limit of the change in y values ((f(x+h)- f(x)), divided by the change in x values (x +h - x= h)

as the changes (the h value) approach zero the difference quotient is the slope of the tangent line at that point. That's the derivative of the function, f(x). The derivative is the slope of a tangent line at a point.

To have a derivative requires continuity. so II is also true. the limit is the same for h approaching zero from the left and right

if the derivative = 0 or is undefined it's a critical value, so also III. if the derivative = 0 that point is also a relative extremum, a local maximum or minimum point. at x=3, f(x) is either a local max or min.

take a specific example

f(x) = y = x^2 - 6x

f'(x) = 2x - 6= the derivative

f'(x) = 2x-6 = 0

2x = 6

x =6/2 = 3

f'(3) =2(3) - 6 = 0 when f(x) =9 -18 = -9 = local minimum = a critical value,

which also happens to be the global minimum

vertex = (3,-9)

[f(3+h) - f(3)]/(x+h -x) = [9+ 6h + h^2 -6h - 18 -9 +18)/h = h^2/h = h

as h approaches 0, [f(3+h) - f(3)]/h approaches 0

the derivative of the function is a critical value

this particular function is an upward opening parabola, a quadratic polynomial.

Had it been a cubic or higher degreed poloynomial, the local extrema wold not be global extrema