Question

The left-hand and right-hand derivatives of f at a are defined byf′−(a)=limh→0−f(a+h)−f(a)h�′−(�)=limℎ→0−�(�+ℎ)−�(�)ℎand f′+(a)=limh→0+f(a+h)−f(a)h�′+(�)=limℎ→0+�(�+ℎ)−�(�)ℎif these limits exist. Then f'(a) exists if and only if these one-sided derivatives exist and are equal.(a) Find f' ^- (4) and f' ^+ (4) for the functionf(x)=⎧⎪⎨⎪⎩0 if x⩽05−x if 0

Answer 1

Explanation:

To find the left-hand derivative of f at x = 4, we need to evaluate:

f′−(4) = limh→0−f(4+h)−f(4)h

Since f(x) = 0 for x ≤ 0 and f(x) = 5 - x for 0 < x < 5, we have:

f(4 + h) = 0 for h < -4

f(4 + h) = 5 - (4 + h) = 1 - h for -4 < h < 1

f(4 + h) = undefined for h > 1

Therefore, we can rewrite the limit as:

f′−(4) = limh→0−f(4+h)−f(4)h = limh→0−(1 - h) - 0h = -1

To find the right-hand derivative of f at x = 4, we need to evaluate:

f′+(4) = limh→0+f(4+h)−f(4)h

Using the same reasoning as before, we can rewrite the limit as:

f′+(4) = limh→0+(5 - (4 + h)) - 0h = -1

Since the left-hand derivative and the right-hand derivative are equal, we can conclude that f'(4) exists and is equal to:

f'(4) = f′−(4) = f′+(4) = -1

Therefore, the derivative of f at x = 4 is -1.