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If f(x)=x(sqrt(x)-sqrt(x+1)) then f(x) i...

If `f(x)=x(sqrt(x)-sqrt(x+1))` then f(x) is:

A

A) Continuous but not differentiable at x = 0

B

B) Differentiable at x = 0

C

C) Not continuous at x = 0

D

D) None of the above

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The correct Answer is:
To determine whether the function \( f(x) = x(\sqrt{x} - \sqrt{x+1}) \) is continuous or differentiable at the point \( x = 0 \), we will follow these steps: ### Step 1: Evaluate \( f(0) \) First, we need to find the value of the function at \( x = 0 \): \[ f(0) = 0(\sqrt{0} - \sqrt{0+1}) = 0(0 - 1) = 0 \] ### Step 2: Find the Left-Hand Limit as \( x \) Approaches 0 Next, we calculate the left-hand limit as \( x \) approaches 0: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x(\sqrt{x} - \sqrt{x+1}) \] Substituting \( x \) with \( -h \) (where \( h \) is a small positive number): \[ \lim_{h \to 0^+} (-h)(\sqrt{-h} - \sqrt{-h + 1}) \] Since \( \sqrt{-h} \) is not defined for real numbers, we can conclude that the left-hand limit does not exist. ### Step 3: Find the Right-Hand Limit as \( x \) Approaches 0 Now, we calculate the right-hand limit as \( x \) approaches 0: \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x(\sqrt{x} - \sqrt{x+1}) \] Substituting \( x \) with \( h \): \[ \lim_{h \to 0^+} h(\sqrt{h} - \sqrt{h+1}) \] This can be simplified: \[ \sqrt{h} - \sqrt{h+1} = \frac{h - (h + 1)}{\sqrt{h} + \sqrt{h+1}} = \frac{-1}{\sqrt{h} + \sqrt{h+1}} \] Thus, \[ \lim_{h \to 0^+} h \left( \frac{-1}{\sqrt{h} + \sqrt{h+1}} \right) = \lim_{h \to 0^+} \frac{-h}{\sqrt{h} + \sqrt{h+1}} = \frac{-0}{\sqrt{0} + \sqrt{1}} = 0 \] ### Step 4: Compare the Limits and the Function Value Now we compare the left-hand limit, right-hand limit, and the function value at \( x = 0 \): - Left-hand limit: Does not exist - Right-hand limit: 0 - Function value: \( f(0) = 0 \) Since the left-hand limit does not exist, the function is not continuous at \( x = 0 \). ### Conclusion Since \( f(x) \) is not continuous at \( x = 0 \), it cannot be differentiable at that point. Thus, the answer is: **The function \( f(x) \) is discontinuous at \( x = 0 \).**
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