If the maximum value of f(x) is 20, then what is the maximum value of (2x - 10) if f(x) = -4x^2+ bx + c?

The maximum value of (2·x - 10) is 20

Step-by-step explanation:

Given that the maximum value of f(x) = 20

f(x) = -4·x² + bx + c

We are required to find the maximum value of (2·x - 10)

f'(x) = -8·x + b = 0

x = b/8

f''(x) = -8, therefore, f'(x) is the maximum point

20 = -4·x² +8·x²  + c

20 = -4·(b/8)² + b×(b/8) + c

20 = -b²/16 + b²/8 + c

20 = b²/16 + c

f(2·x - 10) = -4·(2·x - 10)² + b·(2·x - 10) + c

f(2·x - 10) = b·(2·x - 10) + c - (16·x²-160·x +400

Differentiating to find the maximum gives;

f'(2·x - 10) = d(b·(2·x - 10) + c - (16·x²-160·x +400)/dx = 2·b -32·x +160 = 0

x = (2·b +160)/32 = 0.0625·b +5

At the maximum point, therefore, we have;

b·(2·(0.0625·b +5) - 10) + c - (16·(0.0625·b +5)²-160·(0.0625·b +5) +400

At the max value of f(2·x - 10) = b²/16 + c

Since b²/16 + c = 20, we have the maximum value of (2·x - 10) = 20.

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