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Math Help - Minimum Value of Function

  1. #1
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    Minimum Value of Function

    Q: Find the minimum value of 5 \mathrm{cosh} x + 3 \mathrm{sinh} x.
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    So I know that to find the minimum value, I have to differentiate the equation. Then make it equal to zero and find the value of x by rearranging the equation to make x the subject so I did that ... but got the wrong answer.
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    Method:
    Let y = 5 \mathrm{cosh} x + 3 \mathrm{sinh} x
    \therefore \frac{\mathrm{d}y}{\mathrm{d}x} = 5 \mathrm{sinh} x + 3 \mathrm{cosh} x
    5 \mathrm{sinh} x + 3 \mathrm{cosh} x = 0
    5(e^x - e^{-x})\frac{1}{2} = -3(e^x + e^{-x})\frac{1}{2}
    5e^x - 5e^{-x} = -3e^x - 3e^{-x}
    8e^x - 2e^x = 0
    8e^{2x} - 2 = 0
    8e^{2x} = 2
    e^{2x} = \frac{1}{4}
    2x = \ln \left( \frac{1}{4} \right)
    x = \frac{1}{2} \ln \left( \frac{1}{4} \right).

    ...That is not the correct answer. The correct answer is 4 so where have I gone wrong? Thanks in advance.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    Solving y\,'(x)=0 doesn't give the minimum value of the function, it gives x such that the minimum value is y(x). Hence, computing y\left(-\frac{1}{2}\ln 4\right) should give you 4. (according to my calculator, that's right )
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  3. #3
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    Quote Originally Posted by Air View Post
    Q: Find the minimum value of 5 \mathrm{cosh} x + 3 \mathrm{sinh} x.
    __________________
    So I know that to find the minimum value, I have to differentiate the equation. Then make it equal to zero and find the value of x by rearranging the equation to make x the subject so I did that ... but got the wrong answer.
    __________________
    Method:
    Let y = 5 \mathrm{cosh} x + 3 \mathrm{sinh} x
    \therefore \frac{\mathrm{d}y}{\mathrm{d}x} = 5 \mathrm{sinh} x + 3 \mathrm{cosh} x
    5 \mathrm{sinh} x + 3 \mathrm{cosh} x = 0
    5(e^x - e^{-x})\frac{1}{2} = -3(e^x + e^{-x})\frac{1}{2}
    5e^x - 5e^{-x} = -3e^x - 3e^{-x}
    8e^x - 2e^x = 0
    8e^{2x} - 2 = 0
    8e^{2x} = 2
    e^{2x} = \frac{1}{4}
    2x = \ln \left( \frac{1}{4} \right)
    x = \frac{1}{2} \ln \left( \frac{1}{4} \right).

    ...That is not the correct answer. The correct answer is 4 so where have I gone wrong? Thanks in advance.

    you didn't go wrong, you were asked to find the minimum value of the function you need to finish the question off.

    and also do not use your calculator, it good practise of logs to do this by hand.

    Bobak
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  4. #4
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    How did you understand that it was a function. It didn't say f(x) = \dots?
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  5. #5
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    y is (often) just another name for f(x).
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