# For what value of b does the function 2x^2 + x + 3 and bx^2 - 3x + 1 have the same minimum value.

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### 1 Answer

The value of b has to be determined such that 2x^2 + x + 3 and bx^2 - 3x + 1 have the same minimum value.

For f(x) = 2x^2 + x + 3

f'(x) = 4x + 1

Solving 4x + 1 = 0

=> x = -1/4

Also, f''(x) = 4 which is positive indicating that f(x) has a minimum value at x = -1/4.

At x = -1/4, f(x) = 23/8

For g(x) = bx^2 - 3x + 1

g'(x) = 2bx - 3

Solving 2bx - 3 = 0

=> x = `3/(2b)`

The value of `g(3/(2b))` should be equal to 23/8

=> `b*(3/(2b))^2 - 3*(3/(2b)) + 1 = 23/8`

=> `9/(4*b) - 9/(2*b) = 15/8`

=> `-9/(4*b) = 15/8`

=> b = `(-9*8)/(15*4)`

=> b = -1.2

But g''(x) at -1.2 is negative indicating that the function g(x) has a maximum value at x = -1.2

The graph of f(x) opens upwards while that of g(x) opens downwards. It is only possible to determine the value of b for which the two have the same extreme value.

**At b = -1.2 the given functions have the same extreme value.**