Sorry Guys....I'm completely out of it tonight....

[Daniel_Feldman]

Yep, need help on our calc worksheet. Three problems.

1. Find the minimum value of f(x)=sinhx+2coshx (hyperbolic function). Justify using the second deriivative test. I'm not allowed to use a calculator, or to switch anything to e's.

My solution so far:

f'(x)=coshx+2sinhx
f'(x)=0

coshx+2sinhx=0.

root(1+(sinh^2 x))+2sinhx=0

1+sinh^2 x=4sinh^2 x

3sinh^2 x=1

sinh^2 x-(1/3)=0

(sinhx+(1/root(3)))(sinhx-(1/root(3)))=0

sinhx=1/root(3)
sinhx=-1/root(3)

Here I get stuck. How do I solve this algebraically without using e's?


2. *****SOLVED****** The definite integral, from 0 to x+5 of f(t)dt is equal to x^2+13x+40. Find f(t).


I'm thinking second fundamental theorem, but I'm completely blanking here. If someone can just give me a hint as to the first step, that would be great.


3. Find the definite integral, from -3 to 6, of abs(4-x^2)dx. Note:abs=absolute value.


The integreal comes out to 4x-x^3/3 when x> or equal to 0, and x^3/3-4x when x<0. So the two solutions are 48 and -48. Now i'm just confusing myself, the hint says to graph, which I am doing, and with that I can solve it geometrically, but is there a way this can be done using properties of integration?

 

[galactus]

Yep, need help on our calc worksheet. Three problems.

1. Find the minimum value of f(x)=sinhx+2coshx (hyperbolic function). Justify using the second deriivative test. I'm not allowed to use a calculator, or to switch anything to e's.

My solution so far:

f'(x)=coshx+2sinhx
f'(x)=0

coshx+2sinhx=0.

root(1+(sinh^2 x))+2sinhx=0

1+sinh^2 x=4sinh^2 x

3sinh^2 x=1

sinh^2 x-(1/3)=0

(sinhx+(1/root(3)))(sinhx-(1/root(3)))=0

sinhx=1/root(3)
sinhx=-1/root(3)

Here I get stuck. How do I solve this algebraically without using e's?

You know the minimum is when the derivative is equal to 0,

i.e. cosh(x)+2sinh(x)=0, use Newton's method to arrive at x. I did and got

\(\displaystyle x=-\frac{ln(3)}{2}\)

Therefore, \(\displaystyle sinh(-ln(3)/2)+2cosh(-ln(3)/2)={\sqrt(3)}\)

If you graph sinh(x)+2cosh(x), you will see the minimum at the values above.




2. The definite integral, from 0 to x+5 of f(t)dt is equal to x^2+13x+40. Find f(t).


I'm thinking second fundamental theorem, but I'm completely blanking here. If someone can just give me a hint as to the first step, that would be great.

Hey Dan, I may have something on this one for now. Since we're kind of working backwards here.

Since t=x+5. then x=t-5. Sub into \(\displaystyle x^{2}+13x+40\)

\(\displaystyle (t-5)^{2}+13(t-5)+40=t^{2}+3t\)

Now differentiate. \(\displaystyle (t^{2}+3t)dt=2t+3\)

Now for a check, let's integrate \(\displaystyle 2t+3\), from 0 to x+5.
\(\displaystyle {\int}2t+3dt, t=0..x+5=t^{2}+3t=(x+5)^{2}+3(x+5)=x^{2}+13x+40=(x+5)(x+8)\)

So, \(\displaystyle f(t)=2t+3\)

3. Find the definite integral, from -3 to 6, of abs(4-x^2)dx. Note:abs=absolute value.


The integreal comes out to 4x-x^3/3 when x> or equal to 0, and x^3/3-4x when x<0. So the two solutions are 48 and -48. Now i'm just confusing myself, the hint says to graph, which I am doing, and with that I can solve it geometrically, but is there a way this can be done using properties of integration?

 

[Daniel_Feldman]

Thanks galactus. That look like it works.

 

[soroban]

Hello, Daniel!

Did you try this?

We have: .\(\displaystyle f'(x)\:=\:\cosh x\,+\,2\cdot\sinh x\:=\:0\;\;\Rightarrow\;\;2\cdot\sinh x\:=\:-\cosh x\;\;\Rightarrow\;\;\frac{2\cdot\sinh x}{\cosh x}\:=\:-\frac{1}{2}\)

Hence: .\(\displaystyle \tanh x\:=\:-\frac{1}{2}\)