Three problems I am struggling with, please help!

[Speak_fromtheheart]

So I finally finished my 5 page summer math packet! Phew...but I'm still struggling with a few problems.

1) Evaluate and simplify [f(x + h) - f(x)] / h for f(x) = x^2 - 2x

Maybe I'm just being silly or the notation is weird to me, but can someone just show me how to set it up at least?

2) Simplify e^(1 + ln x)

Ok, this problem just makes no sense to me. I've got it to here -

log e x = 1+ ln x
What do you do from here?

3) sin2x = sinx

I haven't done identities in a looong time. I guess I'm just having trouble with this one. I've got it to here -

2sinxcosx = sinx
Now what?

THANKS! I appreciate it!

 

[stapel]

1) Plug "x + h" in for "x" in f(x). Simplify. Subract f(x) from f(x + h). Simplify. If possible, factor out an "h". Divide the result by "h". If possible, cancel. (You may need to go back and review your notes on "function notation".)

2) I'm sorry, but I don't understand where your equation came from...? Instead, try using the basic properties of logs. For instance, log_b(b^a) = a, and log_b(m) + log_b(n) = log_b(mn). (You may need to go back and review your notes on "logarithmsj" and "log rules".)

3) Get everything over to one side of the equation, equal to zero on the other. Factor the one side, and then set the factors equal to zero. Solve the resulting trig equations over whatever interval was specified. (You may need to go back and review your notes on "trig functions".)

Eliz.

 

[Speak_fromtheheart]

1) Thanks! I think I can do it now.

2) Do you not understand the problem? Or do you not understand how I tried to solve it? I'm pretty good with the basic properties of logs, I'm just stumped on this problem. If you don't understand how I'm trying to solve it, basically I just re wrote the problem as a log.

3) Ok so...

2sinxcosx=sinx
2sinxcosx - sinx = 0
sinx (sinxcosx - 1) = 0
Set the factors equal to zero?

 

[galactus]

#2. Rewrite as \(\displaystyle \L\\e^{1}\cdot{e^{ln(x)}}\)

See now?.

#3. \(\displaystyle \L\\2sin(x)cos(x)=sin(x)\)

Divide by sin(x) and get 2cos(x)=1

Can you finish?.

 

[stapel]

Do you not understand the problem? Or do you not understand how I tried to solve it?

I understand the exercise just fine: that's how I was able to provide you with hints. I said that I didn't understand where you'd gotten the equation from because I didn't see how you were attempting to proceed; that's why I'd asked the question. Sorry for the confusion.

Does "log e x" mean "log-base-e of x", "log-some-other-base of the product of e and x", or something else? What was your reasoning in going from the expression e^{1 + ln(x)} to the equation log e x = 1 + lnx?

Note: If one named the expression, creating the equation y = e^{1 + ln(x)}, then, by taking logs, one would arrive at ln(y) = 1 + ln(x). But that's not what you had...?

#3. \(\displaystyle \L\\2sin(x)cos(x)=sin(x)\)

Divide by sin(x) and get 2cos(x)=1

This assumes that you can divide through by sin(x), which assumes that sin(x) is not zero. Which eliminates some of the solutions to this equation. :shock:

That's why you have to factor:

. . . . .2 sin(x) cos(x) - sin(x) = 0

. . . . .sin(x) [2 cos(x) - 1] = 0

. . . . .sin(x) = 0 or cos(x) = 1/2

Solve both of these to solve the original equation! :wink:

Eliz.

 

[galactus]

Yes, stapel, I failed to mention that.