(b) $$ g(x)=\sqrt[3]{x}-4 x^{5} $$

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It looks like you’re working with the function

g(x) = ∛x – 4x⁵             (where ∛x means the cube root of x).

Could you please clarify what you would like to do with this function? For example, are you interested in:

• Finding its derivative?
 • Determining its domain?
 • Locating critical points or inflection points?
 • Something else?

If you want to, for example, find the derivative of g(x), here is one way to do that:

Step 1. Rewrite the cube root in exponent notation:
  ∛x = x^(1/3).

So the function becomes:
  g(x) = x^(1/3) – 4x⁵.

Step 2. Differentiate term-by-term using the power rule.

• For x^(1/3):
  d/dx [x^(1/3)] = (1/3)x^(1/3 – 1) = (1/3)x^(–2/3).

• For –4x⁵:
  d/dx [–4x⁵] = –4·5x^(5–1) = –20x⁴.

Thus, the derivative is:
  g′(x) = (1/3)x^(–2/3) – 20x⁴.

Note on the Domain of the Derivative:
• The original function g(x) = x^(1/3) – 4x⁵ is defined for all real numbers (since the cube root is defined for all x and polynomials have all of ℝ as their domain).
• However, the expression (1/3)x^(–2/3) is not defined at x = 0 because x^(–2/3) = 1/x^(2/3) and division by zero is not allowed. That means g′(x) is defined for all x ≠ 0.
• In some cases, one can check the derivative at x = 0 by going back to the definition of the derivative. A brief look shows that the limit that defines g′(0) does not settle to a finite value because the difference quotient involves a term behaving like 1/h^(2/3) as h → 0. In other words, the derivative at x = 0 does not exist in the usual (finite) sense.

So, if your task was to find the derivative of g(x), the answer is:

g′(x) = (1/3)x^(–2/3) – 20x⁴      (with x ≠ 0).

Please let me know if you need help with another aspect of this function or further clarification!
The derivative of $$ g(x) = \sqrt[3]{x} - 4x^5 $$ is $$ g'(x) = \frac{1}{3}x^{-\frac{2}{3}} - 20x^4 $$, defined for all $$ x 
eq 0 $$.

(b) \( g(x)=\sqrt[3]{x}-4 x^{5} \)

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