If a function f(x) satisfies the following equation..

Question

If a function $f(x)$ satisfies the following equation $$\int_a^x f(t) \,dt=3x^2+(a+8)x+4$$ then the constant a is [answer 1] and the function $f(x)$ is $f(x)$=[answer 2]. In this obtained function, the minimum of the integral $\int_a^x f(t)dt$ is [answer 3]

I found a through the propertie $\int_a^a f(t)dt=0$ , then $\int_a^a f(t)dt=3a^2+(a+8)a+4=0$ , so a equals -1.

Please, explain how to solve 2 remaining tasks. Thank you for your attention!

Answer 1

By definition, you know that: $$\int_a^x f(t)dt = F(x) - F(a),$$

where

$$\frac{dF(t)}{dt} = f(t).$$

The derivative of $F$ is clearly $6x + a + 8$. Since $a=-1$ ,then:

$$f(t) = 6t + 7.$$

For the third point, you are asked to find the minimum of the function $$3x^2 + 7x + 4.$$

I think that you can solve the third point by yourself.

Answer 2

Part 2

Differentiating both sides

f(x) = 6x + (a+8) = 6x + 7

Part 3

minima occurs if (dy/dx)=0 and (d2y/dx2)>0

So when (dy/dx) = 0, we have 6x+7=0

=> x = -7/6

Now, d2y/dx2 = 6 which is greater than 0, hence it is minima

So value at minima is f(-7/6) = -25/12

Answer 3

put notice that $\int_{a}^{x} f(x) dx = F(x) - F(a) $ when F is a function so that $F' = f$ what means $f(x) = (\int_{a}^{x} f(x) dx)' = 6x+a+8$ and the minimum is achived only if $f(x) = (\int_{a}^{x} f(x) dx)' = 0 \ or \ x = \frac{-(a+8)}{6}$ and also $f(x)' = (\int_{a}^{x} f(x) dx)'' > 0 $ means 6 > 0 which is true.

ans 2 : 6x +a+8

ans 3: $\frac{-(a+8)}{6}$