The sum of all local minimum values of the function f(x) = cases 1 - 2x, & x < -1 (1)/(3)(7 + 2|x|), & -1 ≤ x ≤ 2 (11)/(18)(x - 4)(x - 5), & x > 2 cases$$f(x) = \begin{cases} 1 - 2x, & x < -1 \\ \frac{1}{3}(7 + 2|x|), & -1 \le x \le 2 \\ \frac{11}{18}(x - 4)(x - 5), & x > 2 \end{cases}$$ is:

Related Formula

For a function f(x)$f(x)$ to have a local maximum or minimum at a point, its first derivative must vanish at that point:

f'(x) = 0$f'(x) = 0$

Core Logic

Differentiating the given function f(x) = 6x³ - 45ax² + 108a²x + 1$f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$ with respect to x$x$:

Dividing the entire equation by 18$18$:

Factoring the quadratic equation:

Thus, the critical points are x = 2a$x = 2a$ and x = 3a$x = 3a$. Since a > 0$a > 0$, we assign x₁ = 2a$x_1 = 2a$ and x₂ = 3a$x_2 = 3a$.

Step 1: Finding the value of a

Given that the product of the roots x₁x₂ = 54$x_1x_2 = 54$:

(2a)(3a) = 54$(2a)(3a) = 54$

Since a > 0$a > 0$, we have a = 3$a = 3$.

Step 2: Calculating the final expression

Substituting a = 3$a = 3$ back to find x₁$x_1$ and x₂$x_2$:

x₁ = 2(3) = 6$x_1 = 2(3) = 6$

x₂ = 3(3) = 9$x_2 = 3(3) = 9$

Now, evaluating a + x₁ + x₂$a + x_1 + x_2$:

Pattern Recognition

When critical points are expressed in terms of a parameter, relate the given root condition (x₁x₂ = 54$x_1x_2 = 54$) directly to the product of roots formula ((c)/(a)$\frac{c}{a}$) of the simplified quadratic equation to save factoring time.

Chapter Mix

Class 12 Mathematics: Application of Derivatives

Related Formula

A differentiable function is strictly increasing where its first derivative is positive (f'(x) ≥ 0$f'(x) \ge 0$) and strictly decreasing where its first derivative is negative (g'(x) ≤ 0$g'(x) \le 0$).

Step 1: Differentiate f(x)$f(x)$ to solve for a$a$

Find f'(x)$f'(x)$ :

Since (2,3)$(2,3)$ is the largest open interval of increasing behavior, the transition root occurs at the upper boundary x=3$x=3$ :

Step 2: Differentiate g(x)$g(x)$ to solve for interval (b, c)$(b, c)$

Substitute a = 4$a = 4$ into g(x)$g(x)$:

Compute g'(x)$g'(x)$ using the product rule :

For g(x)$g(x)$ to be strictly decreasing, set g'(x) < 0$g'(x) < 0$ :

Since (x-1)² ≥ 0$(x-1)^2 \ge 0$, we require (x-2)(5x-8) < 0 ⇒ x in ((8)/(5), 2)$(x-2)(5x-8) < 0 \Rightarrow x \in \left(\frac{8}{5}, 2\right)$.

Thus, the interval is (b, c) = ((8)/(5), 2)$(b, c) = \left(\frac{8}{5}, 2\right)$, yielding b = (8)/(5) = 1.6$b = \frac{8}{5} = 1.6$ and c = 2$c = 2$.

Step 3: Compute Target Value

Evaluate the objective expression:

Pattern Recognition

Boundary parameters of maximal monotonic intervals are always the exact zero-crossings of the derivative function. Setting f'(3) = 0$f'(3) = 0$ immediately establishes a=4$a=4$ without secondary algebraic transformations.

Chapter Mix

Class 12 Mathematics: Application of Derivatives

Related Formula

The area of a rectangle bounded between an upper function curve y₂(x)$y_2(x)$ and lower function curve y₁(x)$y_1(x)$ spanning width x = t$x = t$ is formulated as:

Core Logic

The region is bounded below by the line y = x$y = x$ and above by the downward opening parabola y = 9 - (11)/(3)x²$y = 9 - \frac{11}{3}x^2$ in the first quadrant:

Let a vertex of the rectangle lie on the upper parabolic boundary at x = t$x = t$. The corresponding height span of the rectangle is bounded by the line y = t$y = t$ at the base:

Thus, the area equation as a function of variable parameter t$t$ is:

Step 1: Differentiate to Identify Critical Values

Differentiate the area function with respect to t$t$ and equate to zero:

Since x ≥ 0$x \geq 0$, we reject the negative root t = -1$t = -1$. This isolates the physical critical point at:

Step 2: Evaluate Maximum Inscribed Area

Substitute t = (9)/(11)$t = \frac{9}{11}$ back into the factored area equation formulation:

Pattern Recognition

For optimized area allocation inside custom functional borders, setting the optimization parameter strictly to the horizontal coordinates simplifies high-degree polynomials down to standard derivative templates.

Chapter Mix

Class 12 Mathematics: Application of Derivatives

Related Formula

• Functional equations containing x$x$ and (1)/(x)$\frac{1}{x}$ can be solved by substituting x → (1)/(x)$x \to \frac{1}{x}$ to create a system of simultaneous equations.
• A function is strictly increasing in intervals where its first derivative is positive: (dy)/(dx) > 0$\frac{dy}{dx} > 0$.

Core Logic

Given the functional equation equation:

Substitute x → (1)/(x)$x \to \frac{1}{x}$:

Step 1: Solve for f(x)

To eliminate f((1)/(x))$f\left(\frac{1}{x}\right)$, multiply equation (1) by 5 and equation (2) by 4:

Subtract the second equation from the first:

Step 2: Differentiate the Target Equation

We are given y = 9x² f(x) = x² [9f(x)]$y = 9x^2 f(x) = x^2 [9f(x)]$. Substituting our equation for 9f(x)$9f(x)$:

Differentiating with respect to x$x$:

Step 3: Analyze the Monotonicity Intervals

For the function to be strictly increasing, set (dy)/(dx) > 0$\frac{dy}{dx} > 0$:

Using the wavy curve method with critical roots - 1√(5), 0, 1√(5)$-\frac{1}{\sqrt{5}}, \, 0, \, \frac{1}{\sqrt{5}}$:

• Positive regions occur where x in (- 1√(5), 0) ( 1√(5), ∞)$x \in \left(-\frac{1}{\sqrt{5}}, 0\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right)$.

Pattern Recognition

Sees: Symmetric reciprocal arguments inside a functional format. Shortcut: Recognizing that y = 9x² f(x)$y = 9x^2 f(x)$ simplifies directly to an even polynomial (5x⁴ - 2x² - 4$5x^4 - 2x^2 - 4$) means the final derivative will be odd, ensuring a symmetric layout across the origin.

Chapter Mix

Class 12 Mathematics: Application of Derivatives Class 11 Mathematics: Relations and Functions

Related Formula

For local maxima or minima, check the sign flip of f'(x)$f'(x)$ across critical points.

Core Logic

Given f(x) = 2x + 3x2/3$f(x) = 2x + 3x^{2/3}$. Differentiating with respect to x$x$:

Critical points occur where f'(x) = 0$f'(x) = 0$ or where f'(x)$f'(x)$ is undefined:

• f'(x) = 0 x1/3 + 1 = 0 x = -1$f'(x) = 0 \implies x^{1/3} + 1 = 0 \implies x = -1$
• f'(x)$f'(x)$ is undefined at x = 0$x = 0$

Step 1: Sign Scheme Analysis

Let us check the sign changes of f'(x)$f'(x)$:

• For x < -1$x < -1$: x1/3+1 < 0$x^{1/3}+1 < 0$ and x1/3 < 0 f'(x) > 0$x^{1/3} < 0 \implies f'(x) > 0$ (+)
• For -1 < x < 0$-1 < x < 0$: x1/3+1 > 0$x^{1/3}+1 > 0$ and x1/3 < 0 f'(x) < 0$x^{1/3} < 0 \implies f'(x) < 0$ (-)
• For x > 0$x > 0$: x1/3+1 > 0$x^{1/3}+1 > 0$ and x1/3 > 0 f'(x) > 0$x^{1/3} > 0 \implies f'(x) > 0$ (+)

Sign changes:

• At x = -1$x = -1$: positive to negative arrow$\rightarrow$ Local Maxima
• At x = 0$x = 0$: negative to positive arrow$\rightarrow$ Local Minima

Thus, there is exactly one point of local maxima and exactly one point of local minima.

Pattern Recognition

Points where the derivative is undefined (cusps or vertical tangents) are equally valid critical point candidates for local extrema. Always include them in your sign scheme layout.

Chapter Mix

Class 12 Mathematics: Application of Derivatives