Can someone please do these questions step by step and explain them. I really need help so I can do the same for the rest.

a) f(x) = e^(sqrtx)
1. State the domain and range
2. Find f'(x)
3. Determine the intervals of increase and decrease.

d) f(x) = e^x + e^-x
1. State the domain and range
2. Find f'(x)
3. Determine the intervals of increase and decrease.

x must be positive if sqrt x is to be real.

then sqrt x is from 0 to + oo
e^0 = 1
so range would be from 1 to infinity

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

3. f' looks always positive to me

----------------------------
e^x + 1/e^x

think x = - some big number
say -100
then it is e^-100 + e^100
think x the same + big number
then it is e^100 + e^-100
the same y for + 100 and -100 and positive for both

now what if x is + or - 1
then it is e + 1/e around 3

now what if x = 0
1 + 1/1 = 2

so
domain is all real x
range is all y ≥ 2

I think you can do the rest now

a) y = e^(√x)

1. clearly x≥0 for the square root of x to be defined, so the domain is x ≥ 0
range : y ≥ 1
2. y' = (1/2)x^(-1/2)(e^(√x))
= 1/(2√x)(e^√x)
3. the graph is contantly increasing, never decreases

b) this is the equation of a "catenary", or the shape formed when a chain is allowed to hang freely.

it has a vertex when x = 0 giving f(0)= 2. Domain : x any real number, range : y ≥ 2

f'(x) = e^x - e^-x

for x < 0 the function is decreasing (the first derivative is negative)
and for x > 0 the function is increasing, because the derivative is positive.

a) f(x) = e^(sqrt(x))

1. Domain and Range:
To find the domain of a function, we need to check for any restrictions on the input. In this case, the function f(x) = e^(sqrt(x)) involves taking the square root of x. Since the square root is only defined for non-negative numbers, the domain of this function is all x such that x is greater than or equal to 0. So, the domain is [0, +∞).

To determine the range, we need to look at the behavior of the function. The exponential function e^x is always positive, so e^(sqrt(x)) will also be positive for any valid input of x. Therefore, the range of the function is (0, +∞).

2. Finding f'(x):
To find the derivative of f(x), we can use the chain rule. Let's consider the function g(x) = sqrt(x). Applying the chain rule, we have:

f'(x) = (e^(sqrt(x))) * (1 / (2 * sqrt(x)))

So, the derivative of f(x) is f'(x) = (e^(sqrt(x))) / (2 * sqrt(x)).

3. Intervals of Increase and Decrease:
To determine the intervals of increase and decrease for f(x), we need to examine the sign of the derivative, f'(x). Since f'(x) = (e^(sqrt(x))) / (2 * sqrt(x)), we can observe that the denominator is always positive. Therefore, the sign of f'(x) is solely dependent on the exponential term, e^(sqrt(x)).

Since e^(sqrt(x)) is always positive, f'(x) will also be positive for all x in the domain [0, +∞). This means that the function f(x) = e^(sqrt(x)) is always increasing and does not have any intervals of decrease.

d) f(x) = e^x + e^(-x)

1. Domain and Range:
The function f(x) = e^x + e^(-x) does not involve any restrictions on the input x. It is defined for all real numbers. So, the domain of this function is (-∞, +∞).

Regarding the range, we can examine the behavior of the function. The exponential function e^x is always positive, and since e^(-x) is the reciprocal of e^x, it is also positive. Thus, the sum of two positive values will always be greater than 0. Therefore, the range of the function is (0, +∞).

2. Finding f'(x):
To find the derivative of f(x), we need to differentiate each term separately. Applying the derivative rules, we have:

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

So, the derivative of f(x) is f'(x) = e^x - e^(-x).

3. Intervals of Increase and Decrease:
To determine the intervals of increase and decrease for f(x), we need to examine the sign of the derivative, f'(x) = e^x - e^(-x).

The first term, e^x, is always positive. The second term, e^(-x), is the reciprocal of e^x, meaning it is always positive as well.

Since both terms in the derivative are positive, their subtraction will still be positive. This implies that f'(x) = e^x - e^(-x) is positive for all real numbers of x. Therefore, the function f(x) = e^x + e^(-x) is always increasing and does not have any intervals of decrease.