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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