Problem: Let f(x) be a differentiable function such that f(0) = 1 and f'(x) = f(x) + e^x + x^2 - 2, where e is the mathematical constant approximately equal to 2.71828.

a) Find the value of f(1).

b) Find the x-coordinate(s) of the point(s) where the graph of f(x) has a horizontal tangent.

c) Determine the value of the definite integral from 0 to 1 of f(x) with respect to x.

I just came off my vacation and all my math knowledge vanished,i even tried googling it and i couldn't find anything,i guess i am just bad with the internet,although i am a bookworm and i love books i should try the library probably...

Guest Jun 29, 2023

#1**0 **

we'll start by finding the function f(x) that satisfies the given conditions.

To find f(x), let's solve the differential equation f'(x) = f(x) + e^x + x^2 - 2. We can rewrite the equation as:

f'(x) - f(x) = e^x + x^2 - 2.

This is a linear first-order differential equation, and we can solve it using an integrating factor. The integrating factor is given by the exponential of the integral of the coefficient of f(x), which is -1 in this case:

μ(x) = e^(-x).

Multiplying both sides of the differential equation by μ(x), we have:

e^(-x) * [f'(x) - f(x)] = e^(-x) * (e^x + x^2 - 2).

Simplifying the equation, we get:

(e^(-x) * f'(x)) - (e^(-x) * f(x)) = 1 + x^2e^(-x) - 2e^(-x).

Now, notice that the left-hand side is the derivative of (e^(-x) * f(x)) with respect to x. Applying the chain rule, we have:

d/dx (e^(-x) * f(x)) = 1 + x^2e^(-x) - 2e^(-x).

Integrating both sides with respect to x, we obtain:

∫ d/dx (e^(-x) * f(x)) dx = ∫ (1 + x^2e^(-x) - 2e^(-x)) dx.

The integral on the left-hand side can be written as:

e^(-x) * f(x) = ∫ (1 + x^2e^(-x) - 2e^(-x)) dx.

Simplifying the right-hand side, we have:

e^(-x) * f(x) = x - x^2 - 2x + c,

where c is a constant of integration.

To determine the constant c, we use the initial condition f(0) = 1. Plugging in x = 0 and f(x) = 1 into the equation above, we get:

e^(0) * 1 = 0 - 0^2 - 2(0) + c, 1 = c.

Therefore, the equation becomes:

e^(-x) * f(x) = x - x^2 - 2x + 1.

Now, let's solve each part of the problem based on the function we obtained.

a) To find f(1), we substitute x = 1 into the equation:

e^(-1) * f(1) = 1 - 1^2 - 2(1) + 1, e^(-1) * f(1) = -1.

Dividing both sides by e^(-1), we get:

f(1) = -e.

So, the value of f(1) is approximately -2.71828.

b) To find the x-coordinate(s) where the graph of f(x) has a horizontal tangent, we need to find the points where f'(x) = 0.

Given that f'(x) = f(x) + e^x + x^2 - 2, we set f'(x) = 0:

f(x) + e^x + x^2 - 2 = 0.

Unfortunately, finding an analytical solution for this equation is not straightforward. We can use numerical methods such as Newton's method or graphing techniques,also if you have some problems with books and want to write some essays you can access http://supremestudy.com/essay-examples/animal-testing and see diverse essays on different topics from maths to animal testings ! to approximate the x-coordinate(s) where the graph of f(x) has horizontal tangents.

c) To determine the value of the definite integral from 0 to 1 of f(x) with respect to x, we evaluate the integral:

∫[0 to 1] f(x) dx.

We can rewrite the integral using the equation we obtained earlier:

∫[0 to 1] (e^(-x) * f(x)) dx = ∫[0 to 1] (x - x^2 - 2x + 1) dx.

Integrating term by term, we have:

∫[0 to 1] (x - x^2 - 2x + 1) dx = [1/2 * x^2 - 1/3 * x^3 - x^2/2 + x] evaluated from 0 to 1.

Evaluating the integral limits, we get:

[1/2 * (1)^2 - 1/3 * (1)^3 - (1)^2/2 + 1] - [1/2 * (0)^2 - 1/3 * (0)^3 - (0)^2/2 + 0],

[1/2 - 1/3 - 1/2 + 1] - [0],

1/6.

So, the value of the definite integral from 0 to 1 of f(x) with respect to x is 1/6.

hah that took quite a while!

calculatricej Jun 29, 2023