Consider f ' (x) = 6x^2 + 2 and (-1,2) lies on the graph of f. Find the equation for f(x).

Answer 1

Hi Yow S.

f(x) = 2x³ + 2x + 6

Since there is no mention of limits, may I assume you are working with Anti-derivatives please see below

If the first derivative, or f' is 6x² + 2

Then the function can be found by using the antiderivative for a Sum and Power rule as they apply

When the first derivative is axⁿ

The antiderivative is a((xⁿ⁺¹)/(n+ 1))

For 6x²

a=6

n=2

The antiderivative is

6(x²⁺¹)/(2+1) = 6(x³)/3 = 2x³

For 2 remember that anything raised to the 0 power is 1, so 2 is actually

2(x⁰)

For 2x⁰

a=2

n=0

The antiderivative is

2(x⁰⁺¹)/(0+ 1) = 2(x¹)/1 = 2x

To complete your function you need a constant C

Just plug everything back together

f"(x) = 6x² + 2

f(x) = 2x³ + 2x + C

Knowing that your function passes through the point (-1,2), use these coordinates to solve for C.

2 = 2(-1³) + 2(-1) + C

2 = 2(-1) + 2(-1) + C

2 = -2 - 2 + C

2 = -4 + C

Add 4 to both sides of the equation to solve for C

2 + 4 = C

6 = C

So f(x) is

f(x) = 2x³ + 2x + 6

You can check the points by plugging them into the function above.

You can graph your function at Desmos.com