There Is Only One Given Point?

Algebra Level 2

Let f ( x ) = a x 7 + b x 3 + c x − 5 f(x) = ax^7 + bx^3 + cx - 5 for constants a , b a,b and c c . If we are given that f ( − 7 ) = 8 f(-7) = 8 , find f ( 7 ) f(7) .

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Elisabeth Bonnell by Elisabeth Bonnell , 22, USA

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2 solutions

Relevant wiki: Function Graphs

Given: f ( x ) = a x 7 + b x 3 + c x − 5 f(x) = ax^7 + bx^3 + cx - 5 and f ( − 7 ) = 8 f(-7) = 8 .
Consider g ( x ) = a x 7 + b x 3 + c x , g(x) = ax^7 + bx^3 +cx, which is
symmetric across the origin. Thus, f ( x ) f(x) is
five less than g ( x ) g(x) , or g ( x ) = f ( x ) + 5 g(x) = f(x) + 5 and
g ( − 7 ) = 8 + 5 = 13 g(-7) =8 + 5 = 13 . By symmetry across the
origin, g ( 7 ) = − 13 g(7) = -13 . Translating down 5 units,
f ( 7 ) = − 13 − 5 = − 18 f(7) = -13 - 5 = -18 .

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f ( − x ) = − f ( x ) − 10 ; (f(-7) = 8) ⇒ f ( 7 ) = − f ( − 7 ) − 10 = − 8 − 10 = − 18 f(-x) = - f(x) - 10; \quad \text{(f(-7) = 8)} \Rightarrow f(7) = - f(-7) - 10 = -8 - 10 = -18

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