Evaluating an integral without an equation...?

Question

Given:
f is twice differentiable
f(0) = 6
f(1) = 5
f`(1) = 2

1
∫f``(x)dx
0

I'm not sure where to start with this... do I need to figure out an equation or is there a simpler method?

Thus far I'm at 2 - ∫f``(0)dx

Answer 1

My thinking...

Your equation is a second order polynomial,
f(x) = Ax^2 + Bx + C

Since f(0) = 6, C = 6
Since f(1) = 5, A + B = -1

f'(x) = 2Ax + B
f'(1) = 2 = 2A + B
B = 2 - 2A

Putting this back in your f(1) equation,
A + (2 - 2A) = -1

-A = -3
A = 3
B = -4

f(x) = 3x^2 -4x + 6
f(0) = 6
f(1) = 5
f'(x) = 6x - 4 = 2

Now you just need to integrate f(x) = 3x^2 -4x + 6 from 0 to 1.

Hope that helps. :)

Answer 2

let's assume that f(x) is at least a function with a squared term:

f(x) = ax² + bx + c , where a, b, and c are integers.

f'(x) = 2ax + b
...and...
f''(x) = 2a


If you're given f'(1) = 2, try plugging it into our system of equations:

f'(1) = 2a(1) + b = 2a + b = 2

2a = 2 - b
a = (2-b)/(2)

f(0) = 6 = a(0) + b(0) + c , therefore c = 6

f(1) = 5 = a(1²) + b(1) + c = 5

.............. a + b + c = 5
and if c = 6...

a + b + 6 = 5

so a + b = -1

and because we know a = (2-b)/(2)


we can say that

(2-b)/(2) + b = -1

(2-b) + 2b = -2

2 - b + 2b = -2

2 + b = -2
b = -4

recall: a + b = -1, and if b = -4 , a = -1 - (-4) = -1 +4 = 3

therefore,

a = 3, b = -4 , and c = 6

for our function ƒ(x) = 3x² - 4x + 6
we satisfy all the requirements.

I'm not saying this is the ONLY solution, but it is A solution.

Answer 3

The equation f(x) = 3x^2 - 4x +6 fits all your conditions.
f(0) = 6, f(1) = 5, and f'(1) = 2
1
∫f``(x)dx
0
If the above means that you are trying to find f(x) from the second derivative and then evaluate it from 0 to 1, then,

-3(1)^2-4(1) +6 - 0-0-6 = -3 -4 = -7

Answer 4

If f(0)>f(1) and f'(1) > 0, then f' must be negative somewhere between 0 and 1 and have a turning point (or have a vertical asymptote and make the question invalid....).

Okay, that's as far as I got. Sorry.