Let the function $f :[0,5] \to R$ be defined by $f(x)= \begin{cases} 2x+5, & 0\le x1 \\ 2x^2 +5, & 1\le x2 \\ \frac{2}{3} x^3 + \frac{23}{3}, & 2\le x\le5. \end{cases}$ The number of points where $f$ is not differentiable in $(0, 5)$, is __________ .
The question asks for the number of points in the interval $(0, 5)$ where the given piecewise function $f(x)$ is not differentiable.
The function is defined as:
$ f(x)= \begin{cases} 2x+5, & 0\le x \le 1 \\ 2x^2 +5, & 1\le x \le 2 \\ \frac{2}{3} x^3 + \frac{23}{3}, & 2\le x\le5 \end{cases} $We need to check differentiability within the open interval $(0, 5)$. The function is composed of polynomial segments, which are differentiable within their respective open intervals ($0 < x < 1$, $1 < x < 2$, $2 < x < 5$). Therefore, we only need to check the points where the function definition changes: $x=1$ and $x=2$. Both points lie within the interval $(0, 5)$.
Differentiability Check at x=1
- Continuity Check: For differentiability, the function must first be continuous at $x=1$.
- Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2x+5) = 2(1)+5 = 7$.
- Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x^2+5) = 2(1)^2+5 = 7$.
- Function value: $f(1) = 2(1)+5 = 7$.
- Derivative Check: Now, we compare the left-hand and right-hand derivatives.
- Left-hand derivative ($f'_{-}(1)$): For $0 \le x < 1$, $f'(x) = \frac{d}{dx}(2x+5) = 2$. So, $f'_{-}(1) = 2$.
- Right-hand derivative ($f'_{+}(1)$): For $1 < x \le 2$, $f'(x) = \frac{d}{dx}(2x^2+5) = 4x$. So, $f'_{+}(1) = \lim_{x \to 1^+} (4x) = 4(1) = 4$.
Conclusion: $f(x)$ is not differentiable at $x=1$.
Differentiability Check at x=2
- Continuity Check: Check continuity at $x=2$.
- Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x^2+5) = 2(2)^2+5 = 2(4)+5 = 13$.
- Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (\frac{2}{3}x^3 + \frac{23}{3}) = \frac{2}{3}(2)^3 + \frac{23}{3} = \frac{16}{3} + \frac{23}{3} = \frac{39}{3} = 13$.
- Function value: $f(2) = 2(2)^2+5 = 13$.
- Derivative Check: Compare the left-hand and right-hand derivatives.
- Left-hand derivative ($f'_{-}(2)$): For $1 \le x < 2$, $f'(x) = \frac{d}{dx}(2x^2+5) = 4x$. So, $f'_{-}(2) = \lim_{x \to 2^-} (4x) = 4(2) = 8$.
- Right-hand derivative ($f'_{+}(2)$): For $2 < x \le 5$, $f'(x) = \frac{d}{dx}(\frac{2}{3}x^3 + \frac{23}{3}) = \frac{2}{3}(3x^2) = 2x^2$. So, $f'_{+}(2) = \lim_{x \to 2^+} (2x^2) = 2(2)^2 = 8$.
Conclusion: $f(x)$ is differentiable at $x=2$.
Final Count
The function $f(x)$ is not differentiable at $x=1$ within the interval $(0, 5)$. It is differentiable at $x=2$. The function is differentiable within the open intervals $(0,1)$, $(1,2)$, and $(2,5)$.
Therefore, the total number of points where $f$ is not differentiable in $(0, 5)$ is 1.
Important Questions from Differentiability
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