GO Classes DPP | GATE CS, DA | Calculus & Optimization | Differentiability

Answer 1

We are given:

$$F(x) = \max\{x^2 - 6x + 8, \ 2x - 7, \ 1 - x\}$$

Let's define the three individual functions as:

• $f_1(x) = x^2 - 6x + 8$

• $f_2(x) = 2x - 7$

• $f_3(x) = 1 - x$

First, Let's simplify the Function

First, let's see if one of these functions is always less than another. Let's compare the quadratic $f_1(x)$ and the line $f_3(x)$:

$$f_1(x) - f_3(x) = (x^2 - 6x + 8) - (1 - x) = x^2 - 5x + 7$$

To see if this quadratic is always positive, we can check its discriminant ($\Delta = b^2 - 4ac$):

$$\Delta = (-5)^2 - 4(1)(7) = 25 - 28 = -3$$

Since the discriminant is negative and the leading coefficient ($x^2$) is positive, the quadratic $x^2 - 5x + 7$ is strictly greater than $0$ for all real numbers $x$.

This means $f_1(x) > f_3(x)$ everywhere. Therefore, $f_3(x)$ will never be the maximum, and we can drop it from our evaluation:

$$F(x) = \max\{x^2 - 6x + 8, \ 2x - 7\}$$
 

Now we need to find where $f_1(x)$ and $f_2(x)$ intersect to determine which function is the maximum on different intervals.

$$x^2 - 6x + 8 = 2x - 7$$

$$x^2 - 8x + 15 = 0$$

$$(x - 3)(x - 5) = 0$$

The functions intersect at $x = 3$ and $x = 5$.

Testing values in the intervals created by these points, we get the piecewise definition of $F(x)$:

• For $x < 3$: (e.g., $x=0$), $f_1(0) = 8$ and $f_2(0) = -7$. Here, $f_1(x)$ is larger.

• For $3 \le x \le 5$: (e.g., $x=4$), $f_1(4) = 0$ and $f_2(4) = 1$. Here, $f_2(x)$ is larger.

• For $x > 5$: (e.g., $x=6$), $f_1(6) = 8$ and $f_2(6) = 5$. Here, $f_1(x)$ is larger again.

So, the piecewise function is:

$$F(x) = \begin{cases} x^2 - 6x + 8 & \text{if } x < 3 \\ 2x - 7 & \text{if } 3 \le x \le 5 \\ x^2 - 6x + 8 & \text{if } x > 5 \end{cases}$$

 

Let's find the minimum of each piece on its respective interval:

1. Interval $(-\infty, 3]$: The function is a parabola opening upwards. Its vertex is at $x = \frac{-b}{2a} = \frac{6}{2} = 3$. The minimum on this interval is exactly at the boundary $x=3$.

   Value: $F(3) = 3^2 - 6(3) + 8 = -1$.

2. Interval $[3, 5]$: The function is $2x - 7$, which is a strictly increasing line. Its minimum is at the lowest $x$ value, which is $x=3$.

   Value: $F(3) = 2(3) - 7 = -1$.

3. Interval $[5, \infty)$: The function is the parabola $x^2 - 6x + 8$. Because its vertex is at $x=3$, it is strictly increasing for all $x \ge 5$. Its minimum on this interval is at $x=5$.

   Value: $F(5) = 5^2 - 6(5) + 8 = 3$.

Comparing these, the global minimum value is $-1$, and it occurs exactly at $x = 3$.

 

A piecewise function made of polynomials is differentiable everywhere except possibly at the "corner" points where it changes definitions ($x=3$ and $x=5$). We must check the left and right derivatives at these points.

• At $x = 3$:

  • Left derivative: $\frac{d}{dx}(x^2 - 6x + 8)$ evaluated at $x=3 \Rightarrow 2(3) - 6 = \mathbf{0}$

  • Right derivative: $\frac{d}{dx}(2x - 7)$ evaluated at $x=3 \Rightarrow \mathbf{2}$

  • Because $0 \neq 2$, $F(x)$ is not differentiable at $x = 3$.

• At $x = 5$:

  • Left derivative: $\frac{d}{dx}(2x - 7)$ evaluated at $x=5 \Rightarrow \mathbf{2}$

  • Right derivative: $\frac{d}{dx}(x^2 - 6x + 8)$ evaluated at $x=5 \Rightarrow 2(5) - 6 = \mathbf{4}$

  • Because $2 \neq 4$, $F(x)$ is not differentiable at $x = 5$.

 

Now, Let's evaluate the given statements based on our findings:

• A. $F$ is differentiable at its global minimum.

  • False. The global minimum is at $x=3$, and we proved it is not differentiable there.

• B. The minimum value of $F$ is $-1$.

  • True.

• C. The global minimum occurs at $x = 3$.

  • True.

• D. $F$ is not differentiable at exactly two points.

  • True. (It is not differentiable at $x=3$ and $x=5$).