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Zorluk: OrtaFunctions and Custom Symbol Operations

For all real numbers xx, the function gg is defined by g(x)=cx+5g(x) = cx + 5, where cc is a constant. The custom operation ⋄\diamond is defined for all real numbers aa and bb by a⋄b=g(a+b)−g(a−b)a \diamond b = g(a+b) - g(a-b). If 3⋄1=123 \diamond 1 = 12, what is the value of g(4)g(4)?

Cevap: 29

Cevap

29
Applying the function definition g(x)=cx+5g(x) = cx + 5 to the custom operation yields a⋄b=[c(a+b)+5]−[c(a−b)+5]=2bca \diamond b = [c(a+b)+5] - [c(a-b)+5] = 2bc. Substituting a=3a=3 and b=1b=1 into 3⋄1=123 \diamond 1 = 12 gives 2(1)c=122(1)c = 12, so c=6c = 6. Consequently, g(x)=6x+5g(x) = 6x + 5, and evaluating at x=4x = 4 yields g(4)=6(4)+5=29g(4) = 6(4) + 5 = 29.

Adım Adım Çözüm

1
Substitute (a+b)(a+b) and (a−b)(a-b) into the function definition g(x)=cx+5g(x) = cx + 5 to simplify a⋄ba \diamond b.
a⋄b=[c(a+b)+5]−[c(a−b)+5]=2bca \diamond b = [c(a+b) + 5] - [c(a-b) + 5] = 2bc.
Applying the definition of the custom binary operation in terms of function gg eliminates the constant term 55.
2
Use the given equality 3⋄1=123 \diamond 1 = 12 to determine the constant cc.
2(1)c=12  ⟹  2c=12  ⟹  c=62(1)c = 12 \implies 2c = 12 \implies c = 6.
Plugging a=3a=3 and b=1b=1 into 2bc=122bc = 12 yields an equation in terms of cc.
3
Evaluate g(4)g(4) using c=6c = 6.
g(4)=6(4)+5=29g(4) = 6(4) + 5 = 29.
Substituting x=4x = 4 into g(x)=6x+5g(x) = 6x + 5 calculates the required numerical value.

Anahtar Kavram

Functions and Custom Symbol Operations
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