Answer:
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.
