For certain curves y= f(x) satisfying `[d^2y]/[dx^2]= 6x-4`, f(x) has local minimum value 5 when x=1. 9. Number of critical point for y=f(x) for x € [0,2] (a) 0 (b)1. c).2 d) 3 10. Global minimum value y = f(x) for x € [0,2] is (a)5 (b)7 (c)8 d) 9 11 Global maximum value of y = f(x) for x € [0,2] is (a) 5 (b) 7 (c) 8 (d) 9

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the given problem step by step, we will follow the instructions provided in the video transcript.

### Step 1: Start with the given second derivative
We are given the equation:
\[ \frac{d^2y}{dx^2} = 6x - 4 \]

### Step 2: Integrate to find the first derivative
Integrate both sides with respect to \(x\):
\[ \frac{dy}{dx} = \int (6x - 4) \, dx = 3x^2 - 4x + C_1 \]
where \(C_1\) is a constant of integration.

### Step 3: Use the local minimum condition
We know that \(f(x)\) has a local minimum value of 5 when \(x = 1\). At a local minimum, the first derivative is zero:
\[ \frac{dy}{dx} \bigg|_{x=1} = 0 \]
Substituting \(x = 1\) into the first derivative:
\[ 0 = 3(1)^2 - 4(1) + C_1 \implies 0 = 3 - 4 + C_1 \implies C_1 = 1 \]

### Step 4: Write the first derivative with the constant
Now we can write the first derivative:
\[ \frac{dy}{dx} = 3x^2 - 4x + 1 \]

### Step 5: Find critical points
To find the critical points, set the first derivative to zero:
\[ 3x^2 - 4x + 1 = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6} \]
This gives us two solutions:
\[ x = \frac{6}{6} = 1 \quad \text{and} \quad x = \frac{2}{6} = \frac{1}{3} \]
Thus, the critical points are \(x = 1\) and \(x = \frac{1}{3}\).

### Step 6: Determine the nature of critical points
To determine whether these points are maxima or minima, we can check the second derivative:
\[ \frac{d^2y}{dx^2} = 6x - 4 \]
- At \(x = 1\):
\[ \frac{d^2y}{dx^2} \bigg|_{x=1} = 6(1) - 4 = 2 \quad (\text{positive, hence local minimum}) \]
- At \(x = \frac{1}{3}\):
\[ \frac{d^2y}{dx^2} \bigg|_{x=\frac{1}{3}} = 6\left(\frac{1}{3}\right) - 4 = 2 - 4 = -2 \quad (\text{negative, hence local maximum}) \]

### Step 7: Integrate to find the original function
Now integrate the first derivative to find \(y\):
\[ y = \int (3x^2 - 4x + 1) \, dx = x^3 - 2x^2 + x + C_2 \]

### Step 8: Use the local minimum condition to find \(C_2\)
We know that when \(x = 1\), \(y = 5\):
\[ 5 = (1)^3 - 2(1)^2 + (1) + C_2 \implies 5 = 1 - 2 + 1 + C_2 \implies C_2 = 5 \]
Thus, the function is:
\[ y = x^3 - 2x^2 + x + 5 \]

### Step 9: Evaluate \(y\) at the endpoints and critical points
Now we evaluate \(y\) at \(x = 0\), \(x = 1\), \(x = \frac{1}{3}\), and \(x = 2\):
- At \(x = 0\):
\[ y(0) = 0^3 - 2(0)^2 + 0 + 5 = 5 \]
- At \(x = 1\):
\[ y(1) = 1^3 - 2(1)^2 + 1 + 5 = 5 \]
- At \(x = \frac{1}{3}\):
\[ y\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - 2\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right) + 5 = \frac{1}{27} - \frac{2}{9} + \frac{1}{3} + 5 = \frac{1}{27} - \frac{6}{27} + \frac{9}{27} + 5 = \frac{4}{27} + 5 = \frac{139}{27} \approx 5.15 \]
- At \(x = 2\):
\[ y(2) = 2^3 - 2(2)^2 + 2 + 5 = 8 - 8 + 2 + 5 = 7 \]

### Step 10: Determine global minimum and maximum
From the evaluations:
- Global minimum value is \(5\) (occurs at \(x = 0\) and \(x = 1\)).
- Global maximum value is \(y(2) = 7\).

### Summary of Results
1. Number of critical points for \(y = f(x)\) in \([0, 2]\): **2**
2. Global minimum value of \(y = f(x)\) in \([0, 2]\): **5**
3. Global maximum value of \(y = f(x)\) in \([0, 2]\): **7**