Posted by **Sam** on Thursday, June 5, 2008 at 9:50pm.

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.

- Calculus - please help -
**Damon**, Thursday, June 5, 2008 at 10:16pm
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

- Calculus - please help -
**Reiny**, Thursday, June 5, 2008 at 10:22pm
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.

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