Alternative form of derivative Consider the derivative form above.

Alternative form of derivative Consider the derivative form above. Use the derivatives from the left and the right to find the derivative at x = -8 if it exists. (15 points) f(x) = |x + 8

The Correct Answer and Explanation is:

To determine whether the derivative of f(x)=∣x+8∣f(x) = |x + 8| exists at x=−8x = -8, we must evaluate the left-hand and right-hand limits using the alternative form of the derivative:

f′(c)=lim⁡x→cf(x)−f(c)x−cf'(c) = \lim_{{x \to c}} \frac{{f(x) – f(c)}}{{x – c}}

Let c=−8c = -8. Then f(−8)=∣−8+8∣=∣0∣=0f(-8) = | -8 + 8 | = |0| = 0.

Left-hand derivative (as x→−8−x \to -8^-):

When x<−8x < -8, x+8<0x + 8 < 0, so f(x)=−(x+8)f(x) = -(x + 8). Thus,

lim⁡x→−8−f(x)−f(−8)x+8=lim⁡x→−8−−(x+8)−0x+8=lim⁡x→−8−−(x+8)x+8=−1\lim_{{x \to -8^-}} \frac{{f(x) – f(-8)}}{{x + 8}} = \lim_{{x \to -8^-}} \frac{{-(x + 8) – 0}}{{x + 8}} = \lim_{{x \to -8^-}} \frac{{-(x + 8)}}{{x + 8}} = -1

Right-hand derivative (as x→−8+x \to -8^+):

When x>−8x > -8, x+8>0x + 8 > 0, so f(x)=x+8f(x) = x + 8. Thus,

lim⁡x→−8+f(x)−f(−8)x+8=lim⁡x→−8+x+8−0x+8=lim⁡x→−8+x+8x+8=1\lim_{{x \to -8^+}} \frac{{f(x) – f(-8)}}{{x + 8}} = \lim_{{x \to -8^+}} \frac{{x + 8 – 0}}{{x + 8}} = \lim_{{x \to -8^+}} \frac{{x + 8}}{{x + 8}} = 1

Conclusion:

The left-hand derivative is -1 and the right-hand derivative is 1. Since these two one-sided limits are not equal, the derivative at x=−8x = -8 does not exist.

This result is consistent with the nature of absolute value functions. The graph of f(x)=∣x+8∣f(x) = |x + 8| has a sharp corner at x=−8x = -8, which means the slope changes abruptly. A function is not differentiable at a point where it has a corner or cusp because the limit defining the derivative does not approach the same value from both sides. Therefore, the derivative of f(x)f(x) at x=−8x = -8 does not exist.