Find a function whose derivative is Y=(x=1)(x-2) whose graph passes through the point (0,√7)?

Question

Find a function whose derivative is Y=(x=1)(x-2) whose graph passes through the point (0,√7), any help would be appreciated!

Answer 1

I'm going to assume you meant to type + instead of =

So the function in question is

y = (x+1)(x-2). We're going to call this f(x), and we're going to call the antiderivative F(x).

f(x) = (x+1)(x-2)

Your first step would be to expand this.

f(x) = x^2 - x + 1

Now, take the antiderivative (since we want to find the function F(x) with those properties). Taking the antiderivative is simple because they're just powers.

F(x) = (x^3)/3 - (x^2)/2 + x + C

IMPORTANT: Don't forget to add C when taking the antiderivative! C means some constant.

Are we done? Not at all, because the graph must pass through (0, sqrt(7)). This means that we know F(0) = sqrt(7). But

F(0) = (0^3)/3 + (0^2)/2 + 0 + C
So we can substitute F(0) accordingly to get
sqrt(7) = 0 + 0 + 0 + C
C = sqrt(7)

Therefore,

F(x) = (x^3)/3 - (x^2)/2 + x + sqrt(7)

Answer 2

to locate the function f(x), you will could combine the function f'(x) f'(x) = x² + 2 f(x) = x³/3 + 2x + C, the place C is a persevering with. you could evaluate the consistent C from the given archives that the curve passes with the aid of factor (3,40-one) f(3) = (3)³/3 + 2(3) + C = 40-one 9 + 6 + C = 40-one C = 26 consequently, the function you're finding for is: f(x) = x³ + 2x + 26 wish this helps =)

Answer 3

y' = x^2 -x -2 (I think? you had a typo in your question)

y = (1/3)x^3 - (1/2)x^2 - 2x + C

For the point (0, sqrt(7)):
sqrt(7) = (1/3)0^3 - (1/2)(0)^2 - 2(0) + C
C = sqrt(7)

Therefore:

y = (1/3)x^3 - (1/2)x^2 - 2x + sqrt(7)