It's so simple to find the range. Ain't it?
Let f ( x ) = ( x + 1 ) ( x + 2 ) ( x + 3 ) ( x + 4 ) + 5 , where x ∈ [ − 6 , 6 ] .
If the range of the function is [ a , b ] , what is the value of b − a ?
The answer is 5041.
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3 solutions
Transform the equation by putting: [ x = X - 5/2 ] , as the given equation is symmetric. So the function becomes f(x) = (X² - 9/4)(X²- 1/4) + 5 , where [ -7/2 ≤ X ≤ 17/2 ]. From this we can easily get the range.
L e t t = ( X + 1 ) ∗ ( X + 5 ) = X 2 + 5 X + 4 , ∴ t + 2 = ( X + 2 ) ∗ ( X + 3 ) . ∴ f ( t ) = t ∗ ( t + 2 ) + 5 = t 2 + 2 t + 5 . a n d f ′ ( t ) = 2 t + 2 = 0 , ⟹ t = − t f o r e x t r e m u m . F o r e x t r e m u m X 2 + 5 X + 4 = − 1 . S o s o l v i n g t h e q u a d r a t i c w e g e t X = − 1 . 3 8 1 9 7 a n d − 3 . 6 1 8 0 3 . f ( − 1 . 3 8 1 9 7 ) = 4 , a n d f ( − 3 . 6 1 8 0 3 ) = 4 . ∴ b = 4 . I t i s a 4 t h d e g r e e p o l y n o m i a l o p e n i n g u p . ⟹ f ( − 6 ) o r f ( 6 ) i s m a x i m u m . f ( 6 ) > f ( − 6 ) . ∴ a = f ( 6 ) = 5 0 4 5 . a − b = 5 0 4 5 − 4 = 5 0 4 1
Multiply first and third bracket and put x^2+5x as A. Now get a quadratic in A . Its minimum value is -D/4a = 4 . At A=-5 . Hence x lies within interval . One can se maximum value of function will be at x=6 .f(x)=5045