Question

If x ∈ R Let f: R→R be a function given by f(x) = 8x - x² - 4 ,let $\phi$(x) be a function defined by

$\phi$(x) = $\begin{cases} max f(t): x+1 \le t \le x + 2, 0 \le x < 5 \\ min x + 5: 5 \le x \le 8 \end{cases}$,

then $\phi$(x) is

• A. discontinuous and not differentiable
• B. continuous but not differentiable in (0,8)
• C. continous and differentiable in (0,3) $\cup$ (5,$\infty$)
• D. Differentiable at x = 3

Answer 1

Answer: (A)

discontinuous and not differentiable

Answer 2

The function $f(t) = 8t - t^2 - 4 = -(t^2 - 8t + 4) = -(t-4)^2 + 12$. This is a downward-opening parabola with vertex at $t=4$.

For $0 \le x < 5$, $\phi(x) = \max_{t \in [x+1, x+2]} f(t)$. The maximum of $f(t)$ on $[x+1, x+2]$ occurs at the vertex $t=4$ if $x+1 \le 4 \le x+2$ (i.e., $2 \le x \le 3$). In this case, $\phi(x) = f(4) = 12$. If $x+2 \le 4$ (i.e., $x \le 2$), the interval is to the left of the vertex, and $f(t)$ is increasing. The maximum occurs at $t=x+2$. For $0 \le x \le 2$, $\phi(x) = f(x+2) = 8(x+2) - (x+2)^2 - 4 = 8x+16 - (x^2+4x+4) - 4 = -x^2 + 4x + 8$. If $x+1 \ge 4$ (i.e., $x \ge 3$), the interval is to the right of the vertex, and $f(t)$ is decreasing. The maximum occurs at $t=x+1$. For $3 \le x < 5$, $\phi(x) = f(x+1) = 8(x+1) - (x+1)^2 - 4 = 8x+8 - (x^2+2x+1) - 4 = -x^2 + 6x + 3$.

For $5 \le x \le 8$, $\phi(x) = \min \{x+5\}$. Assuming this means $\phi(x) = x+5$ for $5 \le x \le 8$.

So, the function is $\phi(x) = \begin{cases} -x^2 + 4x + 8 & 0 \le x \le 2 \\ 12 & 2 \le x \le 3 \\ -x^2 + 6x + 3 & 3 \le x < 5 \\ x+5 & 5 \le x \le 8 \end{cases}$

Let's check continuity at the transition points $x=2, 3, 5$. At $x=2$: $\lim_{x \to 2^-} \phi(x) = \lim_{x \to 2^-} (-x^2 + 4x + 8) = -(2)^2 + 4(2) + 8 = -4 + 8 + 8 = 12$. $\lim_{x \to 2^+} \phi(x) = \lim_{x \to 2^+} 12 = 12$. $\phi(2) = -(2)^2 + 4(2) + 8 = 12$. Since the limits and function value are equal, $\phi(x)$ is continuous at $x=2$.

At $x=3$: $\lim_{x \to 3^-} \phi(x) = \lim_{x \to 3^-} 12 = 12$. $\lim_{x \to 3^+} \phi(x) = \lim_{x \to 3^+} (-x^2 + 6x + 3) = -(3)^2 + 6(3) + 3 = -9 + 18 + 3 = 12$. $\phi(3) = 12$. Since the limits and function value are equal, $\phi(x)$ is continuous at $x=3$.

At $x=5$: $\lim_{x \to 5^-} \phi(x) = \lim_{x \to 5^-} (-x^2 + 6x + 3) = -(5)^2 + 6(5) + 3 = -25 + 30 + 3 = 8$. $\lim_{x \to 5^+} \phi(x) = \lim_{x \to 5^+} (x+5) = 5 + 5 = 10$. $\phi(5) = 5 + 5 = 10$. Since the left and right limits are not equal, $\phi(x)$ is discontinuous at $x=5$.

Since $\phi(x)$ is discontinuous at $x=5$, it is not differentiable at $x=5$. Thus, $\phi(x)$ is discontinuous in the interval $(0, 8)$. Specifically, it is discontinuous at $x=5$.

Let's check differentiability in $(0, 8)$. For $0 < x < 2$, $\phi'(x) = \frac{d}{dx}(-x^2 + 4x + 8) = -2x + 4$. For $2 < x < 3$, $\phi'(x) = \frac{d}{dx}(12) = 0$. For $3 < x < 5$, $\phi'(x) = \frac{d}{dx}(-x^2 + 6x + 3) = -2x + 6$. For $5 < x < 8$, $\phi'(x) = \frac{d}{dx}(x+5) = 1$.

At $x=2$: Left derivative: $\lim_{x \to 2^-} \phi'(x) = \lim_{x \to 2^-} (-2x + 4) = -2(2) + 4 = 0$. Right derivative: $\lim_{x \to 2^+} \phi'(x) = \lim_{x \to 2^+} 0 = 0$. Since the left and right derivatives are equal, $\phi(x)$ is differentiable at $x=2$ and $\phi'(2) = 0$.

At $x=3$: Left derivative: $\lim_{x \to 3^-} \phi'(x) = \lim_{x \to 3^-} 0 = 0$. Right derivative: $\lim_{x \to 3^+} \phi'(x) = \lim_{x \to 3^+} (-2x + 6) = -2(3) + 6 = 0$. Since the left and right derivatives are equal, $\phi(x)$ is differentiable at $x=3$ and $\phi'(3) = 0$.

At $x=5$: Since $\phi(x)$ is discontinuous at $x=5$, it is not differentiable at $x=5$.

So, $\phi(x)$ is continuous in $(0, 5) \cup (5, 8)$ and differentiable in $(0, 2) \cup (2, 3) \cup (3, 5) \cup (5, 8)$. We found differentiability at $x=2$ and $x=3$. So, $\phi(x)$ is differentiable in $(0, 5)$. However, $\phi(x)$ is discontinuous at $x=5$, so it is not differentiable at $x=5$. Therefore, $\phi(x)$ is discontinuous and not differentiable in $(0, 8)$.