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March 25, 2017

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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 - ,

    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 - ,

    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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