Question 25
A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f (x) at x = 10?
Solution
Let the function be $$ax^2 + bx + c$$.
We know that x=0 value is 1 so c=1.
So equation is $$ax^2 + bx + 1$$.
Now max value is 3 at x = 1.
So after substituting we get a + b = 2.
If f(x) attains a maximum at 'a' then the differential of f(x) at x=a, that is, f'(a)=0.
So in this question f'(1)=0
=> 2*(1)*a+b = 0
=> 2a+b = 0.
Solving the equations we get a=-2 and b=4.
$$ -2x^2 + 4x + 1$$ is the equation and on substituting x=10, we get -159.
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