f(x)=e^x-x

(a) at what x-values does f have relative extrema?
(b) find the exact value of -1 to 1 on an interval f(x)dx.
(c) show that f does not have any inflection points.

f'(x) = e^x - 1

= 0 for extrema of the function
e^x = 1
x = 0

b) ??

c) for points of inflection, f''(x) = 0
f''(x) = e^x
e^x = 0
has no solution.

look at the graph of y = e^x, it never touches the x-axis

(a) To find the x-values where the function f(x)=e^x-x has relative extrema, we need to find the critical points of the function.

To find the critical points, we need to find where the derivative of f(x) equals zero or is undefined. Let's differentiate f(x) with respect to x:

f'(x) = d/dx (e^x - x)

Using the differentiation rules, the derivative of e^x is simply e^x, and the derivative of -x is -1. Therefore, the derivative of f(x) is:

f'(x) = e^x - 1

Now, to find the critical points, we set f'(x) equal to zero and solve for x:

e^x - 1 = 0

Adding 1 to both sides, we get:

e^x = 1

Taking the natural logarithm of both sides, we get:

ln(e^x) = ln(1)
x = 0

So, the only critical point of f(x) is x = 0. Therefore, it is the only potential relative extremum of the function.

(b) To find the exact value of the integral of f(x)dx from -1 to 1, we can evaluate the definite integral. Let's denote the integral as I:

I = ∫[from -1 to 1] (e^x - x) dx

To evaluate the integral, we can antidifferentiate each term separately. The integral of e^x is simply e^x, and the integral of -x is -0.5x^2. Let's evaluate the integral:

I = [e^x - 0.5x^2] from -1 to 1
= (e^1 - 0.5(1)^2) - (e^-1 - 0.5(-1)^2)
= (e - 0.5) - (e^-1 - 0.5)

This gives us the exact value of the integral of f(x)dx from -1 to 1.

(c) To determine whether f(x) has any inflection points, we need to examine the concavity of the function. To do so, we need to find the second derivative of f(x) and analyze its sign changes.

First, let's find the second derivative by differentiating f'(x):

f''(x) = d/dx (e^x - 1)
= e^x

The second derivative of f(x) is e^x. Since e^x is always positive for all real values of x, it never changes sign.

Therefore, f(x) does not have any inflection points, as the concavity of the function does not change.