Question

Let f be a twice differentiable function defined in [-3, 3] such that f(0) = -4, $f'(3) = 0, f'(-3) = 12$ and $f''(x) \ge -2 \forall x \in [-3, 3]$. If $g(x) = \int_{0}^{x} f(t) dt$ then find maximum value of g(x).

Answer 1

48

Answer 2

To find the maximum value of $g(x) = \int_{0}^{x} f(t) dt$ on the interval $[-3, 3]$, we first need to determine the function $f(x)$.

Step 1: Determine $f'(x)$

We are given $f''(x) \ge -2$ for all $x \in [-3, 3]$. Let's define a new function $H(x) = f'(x) + 2x$. Differentiating $H(x)$ with respect to $x$: $H'(x) = f''(x) + 2$. Since $f''(x) \ge -2$, it follows that $f''(x) + 2 \ge 0$, so $H'(x) \ge 0$. This means $H(x)$ is a non-decreasing function on the interval $[-3, 3]$.

Now, let's evaluate $H(x)$ at the endpoints of the interval using the given conditions: At $x = 3$: $H(3) = f'(3) + 2(3) = 0 + 6 = 6$. At $x = -3$: $H(-3) = f'(-3) + 2(-3) = 12 - 6 = 6$.

Since $H(x)$ is a non-decreasing function and its values at the endpoints $x=-3$ and $x=3$ are both $6$, it implies that $H(x)$ must be constant and equal to $6$ for all $x \in [-3, 3]$. Therefore, $f'(x) + 2x = 6$, which gives us: $f'(x) = 6 - 2x$.

Step 2: Determine $f(x)$

Now, integrate $f'(x)$ to find $f(x)$: $f(x) = \int (6 - 2x) dx = 6x - x^2 + C$. We are given $f(0) = -4$. Using this condition to find the constant $C$: $f(0) = 6(0) - (0)^2 + C = C$. So, $C = -4$. Thus, $f(x) = 6x - x^2 - 4$.

Step 3: Determine $g(x)$

Now, substitute $f(x)$ into the definition of $g(x)$: $g(x) = \int_{0}^{x} f(t) dt = \int_{0}^{x} (6t - t^2 - 4) dt$. Integrate the expression: $g(x) = \left[ 3t^2 - \frac{t^3}{3} - 4t \right]_{0}^{x}$ $g(x) = \left( 3x^2 - \frac{x^3}{3} - 4x \right) - \left( 3(0)^2 - \frac{(0)^3}{3} - 4(0) \right)$ $g(x) = 3x^2 - \frac{x^3}{3} - 4x$.

Step 4: Find the maximum value of $g(x)$

To find the maximum value of $g(x)$ on $[-3, 3]$, we need to evaluate $g(x)$ at its critical points within the interval and at the endpoints of the interval. First, find the critical points by setting $g'(x) = 0$. We know $g'(x) = f(x)$. So, $f(x) = 6x - x^2 - 4 = 0$. Rearranging the quadratic equation: $x^2 - 6x + 4 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(4)}}{2(1)}$ $x = \frac{6 \pm \sqrt{36 - 16}}{2}$ $x = \frac{6 \pm \sqrt{20}}{2}$ $x = \frac{6 \pm 2\sqrt{5}}{2}$ $x = 3 \pm \sqrt{5}$.

Now, check which of these critical points lie within the interval $[-3, 3]$: $x_1 = 3 + \sqrt{5} \approx 3 + 2.236 = 5.236$. This is outside $[-3, 3]$. $x_2 = 3 - \sqrt{5} \approx 3 - 2.236 = 0.764$. This is inside $[-3, 3]$.

To determine if $x = 3 - \sqrt{5}$ is a local maximum or minimum, we use the second derivative test: $g''(x) = f'(x) = 6 - 2x$. At $x = 3 - \sqrt{5}$: $g''(3 - \sqrt{5}) = 6 - 2(3 - \sqrt{5}) = 6 - 6 + 2\sqrt{5} = 2\sqrt{5}$. Since $g''(3 - \sqrt{5}) = 2\sqrt{5} > 0$, the critical point $x = 3 - \sqrt{5}$ corresponds to a local minimum for $g(x)$.

Therefore, the maximum value of $g(x)$ must occur at one of the endpoints of the interval $[-3, 3]$. Let's evaluate $g(x)$ at the endpoints $x = -3$ and $x = 3$: $g(3) = 3(3)^2 - \frac{(3)^3}{3} - 4(3)$ $g(3) = 3(9) - \frac{27}{3} - 12$ $g(3) = 27 - 9 - 12$ $g(3) = 18 - 12 = 6$.

$g(-3) = 3(-3)^2 - \frac{(-3)^3}{3} - 4(-3)$ $g(-3) = 3(9) - \frac{-27}{3} + 12$ $g(-3) = 27 - (-9) + 12$ $g(-3) = 27 + 9 + 12$ $g(-3) = 36 + 12 = 48$.

Comparing the values: $g(0) = 0$, $g(3) = 6$, and $g(-3) = 48$. The maximum value of $g(x)$ on the interval $[-3, 3]$ is $48$.