Pregunta
Select the true statements. \( \square \) A. The function \( f(x)=x^{3} \) is differentiable on \( [0,7] \). \( \square \) B. The function \( f(x)=\sqrt{x} \) is differentiable on \( [0,7] \) \( \square \) C. The function \( f(x)=|x| \) is differentiable on \( (-3,3) \) \( \square \) D. The function \( f(x)=\sqrt{x} \) is differentiable on \( (0,7) \) \( \square \) E. The function \( f(x)=|x| \) is differentiable on \( [0,7] \).
Ask by Huang Martinez. in the United States
Mar 09,2025
Solución de inteligencia artificial de Upstudy
Respuesta verificada por el tutor
Responder
The true statements are A, D, and E.
Solución
1. For statement A, the function is \( f(x)=x^3 \).
- Since polynomials are differentiable everywhere, \( f(x) \) is differentiable on \([0,7]\).
- Statement A is true.
2. For statement B, the function is \( f(x)=\sqrt{x} \).
- Its derivative is \( f'(x)=\frac{1}{2\sqrt{x}} \).
- At \( x=0 \), the derivative would require evaluating \(\frac{1}{2\sqrt{0}}\), which is undefined (or infinite).
- Thus, \( f(x)=\sqrt{x} \) is not differentiable at \( x=0 \), and consequently not differentiable on the entire interval \([0,7]\).
- Statement B is false.
3. For statement C, the function is \( f(x)=|x| \).
- The function \( |x| \) has a derivative of \(-1\) for \( x<0 \) and \( 1 \) for \( x>0 \).
- At \( x=0 \), the left-hand derivative is \(-1\) and the right-hand derivative is \(1\); they are not equal.
- Hence, \( f(x) \) is not differentiable at \( x=0 \), so it is not differentiable on the entire interval \((-3,3)\).
- Statement C is false.
4. For statement D, the function is again \( f(x)=\sqrt{x} \), but now the interval is \((0,7)\).
- On \((0,7)\), all \( x \) are positive, so the derivative \( f'(x)=\frac{1}{2\sqrt{x}} \) exists for every \( x \) in the interval.
- Therefore, \( f(x)=\sqrt{x} \) is differentiable on \((0,7)\).
- Statement D is true.
5. For statement E, the function is \( f(x)=|x| \) on \([0,7]\).
- For \( x>0 \), \( f(x)=|x|=x \) which is differentiable with derivative \( 1 \).
- At the endpoint \( x=0 \) (since this is a closed interval), we only need to check the right-hand derivative.
- The right-hand derivative at \( x=0 \) is
\[
\lim_{h \to 0^+} \frac{|h|-0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1.
\]
- Since the one-sided derivative exists at \( x=0 \) and the function is differentiable for all \( x>0 \), \( f(x)=|x| \) is differentiable on \([0,7]\).
- Statement E is true.
The true statements are A, D, and E.
Respondido por UpStudy AI y revisado por un tutor profesional
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Beyond the Answer
The function \( f(x)=x^{3} \) is indeed differentiable on the interval \( [0,7] \). Polynomial functions like this one are smooth and continuous everywhere, so you're good here! On the flip side, functions like \( f(x)=|x| \) have a "sharp corner" at \( x=0 \), making them non-differentiable at that point. Specifically, it's differentiable on \( (-3,3) \), but not on the closed interval \( [0,7] \) because of that corner at \( x=0 \).

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