Question

determine if the function $f(x)=\sqrt4{x^3}$ is differentiable at $x = 0$. if not, identify why.\
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$\boldsymbol{f(x)=\sqrt4{x^3}}$ is not differentiable at $x = 0$ because $f(0)$ is not defined.\
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$\boldsymbol{f(x)=\sqrt4{x^3}}$ is differentiable at $x = 0$.\
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$\boldsymbol{f(x)=\sqrt4{x^3}}$ is not differentiable at $x = 0$ because $\lim\limits_{x\to 0^+}\sqrt4{x^3}$ does not exist.\
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$\boldsymbol{f(x)=\sqrt4{x^3}}$ is not differentiable at $x = 0$ because $f(0)$ is not defined.

Explanation:

Step1: Rewrite function in exponent form

$f(x) = \sqrt[4]{x^3} = x^{\frac{3}{4}}$

Step2: Check $f(0)$ is defined

$f(0) = 0^{\frac{3}{4}} = 0$, so $f(0)$ exists.

Step3: Find derivative of $f(x)$

Using power rule: $f'(x) = \frac{3}{4}x^{\frac{3}{4}-1} = \frac{3}{4}x^{-\frac{1}{4}} = \frac{3}{4x^{\frac{1}{4}}}$

Step4: Evaluate $f'(0)$

Substitute $x=0$: $f'(0) = \frac{3}{4\cdot0^{\frac{1}{4}}}$, which is undefined (division by 0).

Step5: Analyze limit of $f(x)$ at $0^+$

$\lim_{x \to 0^+} \sqrt[4]{x^3} = 0$, so this limit exists.

Answer:

$f(x)=\sqrt[4]{x^3}$ is not differentiable at $x = 0$ because $f'(0)$ is not defined.