Soru

Zorluk: ZorFunctions and Custom Symbol Operations

For any real number xx, the custom unary operation △\triangle is defined by △(x)=2x−x2\triangle(x) = 2x - x^2. For all real numbers uu and vv, the custom binary operation ⊙\odot is defined by u⊙v=△(u)+△(v)+uvu \odot v = \triangle(u) + \triangle(v) + uv. The function ff is defined for all real numbers xx by f(x)=x⊙(3−x)f(x) = x \odot (3 - x). What is the maximum value of f(x)f(x)?

Cevap: 3.75

Cevap

The maximum value of f(x)f(x) is 3.753.75 (or 154\frac{15}{4}).
Expanding f(x)=x⊙(3−x)f(x) = x \odot (3 - x) yields △(x)+△(3−x)+x(3−x)\triangle(x) + \triangle(3 - x) + x(3 - x). Substituting △(t)=2t−t2\triangle(t) = 2t - t^2 gives (2x−x2)+(6−2x−(9−6x+x2))+(3x−x2)=−3x2+9x−3(2x - x^2) + (6 - 2x - (9 - 6x + x^2)) + (3x - x^2) = -3x^2 + 9x - 3. The vertex of this quadratic function occurs at x=96=1.5x = \frac{9}{6} = 1.5, where f(1.5)=−3(2.25)+13.5−3=3.75f(1.5) = -3(2.25) + 13.5 - 3 = 3.75.

Adım Adım Çözüm

1
Evaluate the custom unary operation △\triangle for each argument
△(x)=2x−x2\triangle(x) = 2x - x^2 and △(3−x)=2(3−x)−(3−x)2=−x2+4x−3\triangle(3-x) = 2(3-x) - (3-x)^2 = -x^2 + 4x - 3
Substitute xx and 3−x3-x into the definition △(t)=2t−t2\triangle(t) = 2t - t^2 and expand carefully.
2
Compute the product term uvuv
x(3 - x) = 3x - x^2
The binary definition u⊙vu \odot v includes an additive product term uvuv.
3
Sum all components to construct the explicit quadratic expression for f(x)f(x)
f(x) = (2x - x^2) + (-x^2 + 4x - 3) + (3x - x^2) = -3x^2 + 9x - 3
Combine like terms for x2x^2, xx, and the constant.
4
Determine the vertex of the downward-opening parabola f(x)=−3x2+9x−3f(x) = -3x^2 + 9x - 3
x = \frac{3}{2} = 1.5 ,yielding, yielding f(1.5) = 3.75$
Since the coefficient of x2x^2 is negative (−3<0-3 < 0), the maximum occurs at x=−b2a=96=1.5x = -\frac{b}{2a} = \frac{9}{6} = 1.5.

Anahtar Kavram

Custom operations combined with quadratic function optimization
Tahmini Süre:2m 0s
Bu soruyu puanla