Minimum Value of Function

Q: Find the minimum value of $\displaystyle 5 \mathrm{cosh} x + 3 \mathrm{sinh} x$.
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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 $\displaystyle x$ by rearranging the equation to make $\displaystyle x$ the subject so I did that ... but got the wrong answer. (Crying)
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Method:
Let $\displaystyle y = 5 \mathrm{cosh} x + 3 \mathrm{sinh} x$
$\displaystyle \therefore \frac{\mathrm{d}y}{\mathrm{d}x} = 5 \mathrm{sinh} x + 3 \mathrm{cosh} x$
$\displaystyle 5 \mathrm{sinh} x + 3 \mathrm{cosh} x = 0$
$\displaystyle 5(e^x - e^{-x})\frac{1}{2} = -3(e^x + e^{-x})\frac{1}{2}$
$\displaystyle 5e^x - 5e^{-x} = -3e^x - 3e^{-x}$
$\displaystyle 8e^x - 2e^x = 0$
$\displaystyle 8e^{2x} - 2 = 0$
$\displaystyle 8e^{2x} = 2$
$\displaystyle e^{2x} = \frac{1}{4}$
$\displaystyle 2x = \ln \left( \frac{1}{4} \right)$
$\displaystyle x = \frac{1}{2} \ln \left( \frac{1}{4} \right)$.

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

Hi

Solving $\displaystyle y\,'(x)=0$ doesn't give the minimum value of the function, it gives $\displaystyle x$ such that the minimum value is $\displaystyle y(x)$. Hence, computing $\displaystyle y\left(-\frac{1}{2}\ln 4\right)$ should give you $\displaystyle 4$. (according to my calculator, that's right (Clapping))

Quote:

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Originally Posted by Air

Q: Find the minimum value of $\displaystyle 5 \mathrm{cosh} x + 3 \mathrm{sinh} x$.
__________________
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 $\displaystyle x$ by rearranging the equation to make $\displaystyle x$ the subject so I did that ... but got the wrong answer. (Crying)
__________________
Method:
Let $\displaystyle y = 5 \mathrm{cosh} x + 3 \mathrm{sinh} x$
$\displaystyle \therefore \frac{\mathrm{d}y}{\mathrm{d}x} = 5 \mathrm{sinh} x + 3 \mathrm{cosh} x$
$\displaystyle 5 \mathrm{sinh} x + 3 \mathrm{cosh} x = 0$
$\displaystyle 5(e^x - e^{-x})\frac{1}{2} = -3(e^x + e^{-x})\frac{1}{2}$
$\displaystyle 5e^x - 5e^{-x} = -3e^x - 3e^{-x}$
$\displaystyle 8e^x - 2e^x = 0$
$\displaystyle 8e^{2x} - 2 = 0$
$\displaystyle 8e^{2x} = 2$
$\displaystyle e^{2x} = \frac{1}{4}$
$\displaystyle 2x = \ln \left( \frac{1}{4} \right)$
$\displaystyle x = \frac{1}{2} \ln \left( \frac{1}{4} \right)$.

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

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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

How did you understand that it was a function. It didn't say $\displaystyle f(x) = \dots$?

$\displaystyle y$ is (often) just another name for $\displaystyle f(x)$.