lim h-0 f(x) = 8 – Ixl Choose the correct answer below: The function is not differentiable at x = 0 because the left- and right-hand limits of the difference quotient exist at x = 0, but are not equal: The function is not differentiable at x = 0 because the left- and right-hand limits of the difference quotient do not exist at x= 0. The function is differentiable at x = 0 because the graph has a sharp corner atx= 0. The function is differentiable at x = 0 because both the left- and right-hand limits of the difference quotient exist at x = 0.
f(0 + h) – f(0) Determine whether is differentiable at x = 0 by considering lim h-0 f(x) = 8 – Ixl Choose the correct answer below: The function is not differentiable at x = 0 because the left- and right-hand limits of the difference quotient exist at x = 0, but are not equal: The function is not differentiable at x = 0 because the left- and right-hand limits of the difference quotient do not exist at x= 0. The function is differentiable at x = 0 because the graph has a sharp corner atx= 0. The function is differentiable at x = 0 because both the left- and right-hand limits of the difference quotient exist at x = 0.
The Correct Answer and Explanation is:
The function in question is f(x)=8−∣x∣f(x) = 8 – |x|f(x)=8−∣x∣, and we are interested in determining its differentiability at x=0x = 0x=0.
Step 1: Calculate the left-hand and right-hand limits of the difference quotient.
To determine whether the function is differentiable at x=0x = 0x=0, we need to check the behavior of the difference quotient as h→0h \to 0h→0. The difference quotient is:f(x+h)−f(x)h\frac{f(x + h) – f(x)}{h}hf(x+h)−f(x)
At x=0x = 0x=0, this becomes:f(0+h)−f(0)h=f(h)−f(0)h\frac{f(0 + h) – f(0)}{h} = \frac{f(h) – f(0)}{h}hf(0+h)−f(0)=hf(h)−f(0)
We know that f(x)=8−∣x∣f(x) = 8 – |x|f(x)=8−∣x∣, so:
- When x>0x > 0x>0, f(x)=8−xf(x) = 8 – xf(x)=8−x.
- When x<0x < 0x<0, f(x)=8+xf(x) = 8 + xf(x)=8+x.
- f(0)=8f(0) = 8f(0)=8 (since f(0)=8−∣0∣=8f(0) = 8 – |0| = 8f(0)=8−∣0∣=8).
For h>0h > 0h>0 (right-hand limit):
f(h)−f(0)h=(8−h)−8h=−hh=−1\frac{f(h) – f(0)}{h} = \frac{(8 – h) – 8}{h} = \frac{-h}{h} = -1hf(h)−f(0)=h(8−h)−8=h−h=−1
For h<0h < 0h<0 (left-hand limit):
f(h)−f(0)h=(8+h)−8h=hh=1\frac{f(h) – f(0)}{h} = \frac{(8 + h) – 8}{h} = \frac{h}{h} = 1hf(h)−f(0)=h(8+h)−8=hh=1
Step 2: Evaluate the limits
- As h→0+h \to 0^+h→0+ (from the right), the limit of the difference quotient is −1-1−1.
- As h→0−h \to 0^-h→0− (from the left), the limit of the difference quotient is 111.
Step 3: Conclusion
Since the left-hand and right-hand limits of the difference quotient are not equal, the function is not differentiable at x=0x = 0x=0.
Correct Answer:
The function is not differentiable at x = 0 because the left- and right-hand limits of the difference quotient exist at x = 0, but are not equal.
Explanation:
The fact that the left-hand and right-hand limits are different (one is −1-1−1 and the other is 111) means that the slope of the function from the left side is different from the slope from the right side at x=0x = 0x=0. This is a clear indicator of a “corner” in the graph of the function at that point, which causes the function to fail the differentiability test at x=0x = 0x=0. Therefore, the function is not differentiable at this point, and this sharp corner behavior is a key feature of functions like f(x)=8−∣x∣f(x) = 8 – |x|f(x)=8−∣x∣.
