For certain curve `y=f(x)` satisfying `(d^(2)y)/(dx^(2))=6x-4, f(x)` has local minimum value 5 when `x=1`
Global maximum value of `y=f(x)` for `x in [0,2]` is

Options

A. 5
B. 7
C. 8
D. 9

Explanation

To solve the problem, we will follow these steps:

1. **Integrate the second derivative**: We start with the given second derivative of the function \( \frac{d^2y}{dx^2} = 6x - 4 \). We will integrate this to find the first derivative \( \frac{dy}{dx} \).

\[ \frac{dy}{dx} = \int (6x - 4) \, dx = 3x^2 - 4x + C_1 \]

Here, \( C_1 \) is a constant of integration.

2. **Use the local minimum condition**: We know that there is a local minimum at \( x = 1 \). For a local minimum, the first derivative must be zero at that point.

\[ 0 = 3(1)^2 - 4(1) + C_1 \]
\[ 0 = 3 - 4 + C_1 \implies C_1 = 1 \]

Thus, the first derivative becomes:

\[ \frac{dy}{dx} = 3x^2 - 4x + 1 \]

3. **Integrate again to find \( y \)**: Now we will integrate \( \frac{dy}{dx} \) to find \( y \).

\[ y = \int (3x^2 - 4x + 1) \, dx = x^3 - 2x^2 + x + C_2 \]

Here, \( C_2 \) is another constant of integration.

4. **Use the local minimum value**: We know that \( f(1) = 5 \). We can use this to find \( C_2 \).

\[ 5 = (1)^3 - 2(1)^2 + (1) + C_2 \]
\[ 5 = 1 - 2 + 1 + C_2 \implies C_2 = 5 \]

Thus, the function \( y \) is:

\[ y = x^3 - 2x^2 + x + 5 \]

5. **Find critical points**: To find the global maximum, we need to find the critical points by setting the first derivative to zero.

\[ 3x^2 - 4x + 1 = 0 \]

We can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6} \]

This gives us:

\[ x = 1 \quad \text{and} \quad x = \frac{1}{3} \]

6. **Evaluate \( y \) at critical points and endpoints**: We will evaluate \( y \) at \( x = 0, \frac{1}{3}, 1, \) and \( 2 \).

- For \( x = 0 \):
\[ y(0) = 0^3 - 2(0)^2 + 0 + 5 = 5 \]

- For \( x = \frac{1}{3} \):
\[ y\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - 2\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right) + 5 = \frac{1}{27} - \frac{2}{9} + \frac{1}{3} + 5 = \frac{1}{27} - \frac{6}{27} + \frac{9}{27} + 5 = \frac{4}{27} + 5 = \frac{139}{27} \]

- For \( x = 1 \):
\[ y(1) = 1^3 - 2(1)^2 + 1 + 5 = 1 - 2 + 1 + 5 = 5 \]

- For \( x = 2 \):
\[ y(2) = 2^3 - 2(2)^2 + 2 + 5 = 8 - 8 + 2 + 5 = 7 \]

7. **Determine the global maximum**: Now we compare the values obtained:

- \( y(0) = 5 \)
- \( y\left(\frac{1}{3}\right) = \frac{139}{27} \approx 5.148 \)
- \( y(1) = 5 \)
- \( y(2) = 7 \)

The global maximum value of \( y \) on the interval \([0, 2]\) is:

\[ \text{Global Maximum} = 7 \]