Minimum Value of Function

• May 7th 2008, 10:17 AM
Simplicity
Minimum Value of Function
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.
• May 7th 2008, 10:38 AM
flyingsquirrel
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))
• May 7th 2008, 10:42 AM
bobak
Quote:

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.

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
• May 7th 2008, 10:46 AM
Simplicity
How did you understand that it was a function. It didn't say $\displaystyle f(x) = \dots$?
• May 7th 2008, 10:52 AM
xifentoozlerix
$\displaystyle y$ is (often) just another name for $\displaystyle f(x)$.