Author Topic: BEC'S methods questions (Read 108216 times) Tweet Share

bec · « **on:** December 30, 2007, 10:48:07 pm »

OK so I've just had a crack at doing a couple of the chapter reviews from the methods textbook and hit a wall at differentiation techniques. Can anyone help with these questions? I might just post one at a time for simplicity's sake (for anyone who has Mathsworld 3/4, it's on page 202)

Let f:R-->R, f(x) = e^2x + 2ke^x + 1, where k is a real number
a) For what values of k will the graph of f have a stationary point?
The way I did it was:
f'(x) = 2e^2x + 2ke^x = 0
2e^2x (1+k) = 0
1+k = 0
k = -1
but according to the answers, it should be "k<0"

?

...and that is where i gave up....

b) Find the coordinates of this stationary point when it exists.
c) Find the value of k for which the graph of f touches the x-axis.

I'd really appreciate it if anyone could point me in the right direction here!

Collin Li · « **Reply #1 on:** December 30, 2007, 10:59:57 pm »

For part a), you factorised it incorrectly.

Here is my solution:

$f(x) = e^{2x} + 2ke^x + 1 \Rightarrow f'(x) = 2e^{2x} + 2ke^x = 2(e^x)^2 + 2ke^x = 2e^x(e^x + k)$

Stationary points: $f'(x) = 0$ :

$\Rightarrow 0 = 2e^x(e^x + k)$

Since $2e^x \neq 0$ : $\Rightarrow 0 = e^x + k \Rightarrow x = \log_e(-k)$

Since $\log_e(-k)$ is only defined for $-k > 0 \Rightarrow k < 0$ (i.e.: if $k$ is not less than zero, there won't be a solution to $f'(x) = 0$ , since the solution is $x = \log_e(-k)$

Collin Li · « **Reply #2 on:** December 30, 2007, 11:07:12 pm »

Co-ordinates of the stationary point

$f(x) = f(\log_e(-k)) = (-k)^2 + 2k\cdot (-k) + 1 = k^2 - 2k^2 + 1 = 1 - k^2$

Therefore, the co-ordinates are: $\left(\log_e(-k),1 - k^2\right)$

Collin Li · « **Reply #3 on:** December 30, 2007, 11:11:15 pm »

Find the value of k for which the graph of f touches the x-axis.

Touching the x-axis requires:
* derivative is zero, and
* function is zero

We know that the derivative is zero for $x = \log_e(-k)$ , and that the function at that point is $1-k^2$ (see previous question).

Therefore we must fix $1-k^2 = 0 \Rightarrow 1 = k^2 \Rightarrow k = \pm 1$

However, since $k < 0$ , as determined in part a):

$\Rightarrow k = -1$

bec · « **Reply #4 on:** December 30, 2007, 11:14:48 pm »

Thank you so much, you are amazing!
It actually seems really clear now which is surprising since normally I can't do maths after 8.30pm...I'll look at it again tomorrow when I'm more awake though to soak it in a bit. Thanks!

bec · « **Reply #5 on:** December 31, 2007, 08:49:42 am »

Another question:

x 1 2 3 4
f(x) 3 2 1 4
f'(x) 1 4 2 3
g(x) 2 1 4 3
g'(x) 4 2 3 1

Using the information about f and g and their derivatives f' and g' in the table above, find:
a) h'(1) if h(x) = f[g(x)]
b) h'(2) if h(x) = g[f(x)]
c) h'(3) if h(x) = f(x)g(x)
d) h'(4) if h(x) = g(x)/f(x)

I don't even know where to start, it's obviously a conceptual gap in my brain...help!

Collin Li · « **Reply #6 on:** December 31, 2007, 11:15:24 am »

Strategy: obtain an expression for $h'(x)$ . We have a rule for $h(x)$ , so we just can derive.

a)
$\mbox{If }h(x) = f(g(x))\mbox{, then }h'(x) = g'(x)\cdot f'(g(x))$ (This is the chain rule)

$\Rightarrow h'(1) = g'(1)\cdot f'(g(1)) = 4\cdot f'(2) = 4\cdot 4 = 16$

b)
$\mbox{If }h(x) = g(f(x))\mbox{, then }h'(x) = f'(x)\cdot g'(f(x))$ (This is the chain rule again)

$\Rightarrow h'(2) = f'(2)\cdot g'(f(2)) = 4\cdot g'(2) = 4\cdot 2 = 8$

c)
$\mbox{If }h(x) = f(x)\cdot g(x)\mbox{, then }h'(x) = f'(x)\cdot g(x) + g'(x)\cdot f(x)$ (This is the product rule)

$\Rightarrow h'(3) = f'(3)\cdot g(3) + g'(3)\cdot f(3) = 2\cdot 4 + 3\cdot 1 = 11$

d)
$\mbox{If }h(x) = \frac{g(x)}{f(x)}\mbox{, then }h'(x) = \frac{g'(x)\cdot f(x) - f'(x)\cdot g(x)}{\left[f(x)\right]^2}$ (This is the quotient rule)

$\Rightarrow h'(4) = \frac{g'(4)\cdot f(4) - f'(4)\cdot g(4)}{\left[f(4)\right]^2} = \frac{1\cdot 4 - 3\cdot 3}{4^2} = -\frac{5}{16}$

bec · « **Reply #7 on:** December 31, 2007, 12:46:20 pm »

arggggh so obvious! thank you so much

i have another question too: it seems easy but i just can't get the correct answer without using my calc to draw a graph...

Consider the curves whose equations are y=cos2x and y=sinx for 0≤x≤π (that's meant to be pi)
Find the coordinates of the points of intersection for these two curves.
Here's what I did...
Point of intersection where cos2x=sinx
By identity, cos2x = 1 - 2sin²x
Therefore: 1 - 2sin²x = sinx
1 - 2sin²x - sinx = 0....This is now a quadradratic. Let a=sinx
1 - 2a²-a = 0
-(a+1)(2a-1) = 0, therefore a=-1 OR 1/2

Sub a=sinx back in.
sinx=-1
x= not defined in this domain

sinx=1/2
x=π/6
When x=π/6, y=1/2....(π/6, 1/2) <---------This one at least is right!

I know the other other solution is when x=5π/6 but I can't figure it out algebraically. What am I doing wrong?

cara.mel · « **Reply #8 on:** December 31, 2007, 12:49:17 pm »

You're going to hate this - you've done everything absolutely right

All the solutions for when sin x = 1/2, 0≤x≤π are pi/6 and 5pi/6. That's it. Well done

bec · « **Reply #9 on:** December 31, 2007, 12:54:33 pm »

haha did i?
But I didn't get the second solution, the 5pi/6 one....the only reason i wrote it in my post was because i was saying, "i know that the other one IS this"...but i can't get it.
how do you find 5pi/6?

(ps whoa that reply was quick!)

lishan515 · « **Reply #10 on:** December 31, 2007, 12:58:09 pm »

quadrant - sinx=1/2 - when positive quadrant

s a - therefore in 1st and 2nd - so pi/6 and 5pi/6
t c

bec · « **Reply #11 on:** December 31, 2007, 01:01:42 pm »

ah thanks lishan515!
i feel like an idiot now. but it's holidays so i'm allowed to...

bec · « **Reply #12 on:** December 31, 2007, 01:55:12 pm »

4 more to go on my i-don't-get-it list...

Consider the family of curves defined by the equation y = xⁿ/(x²+1), where n is a positive integer.
i) For what value(s) of n does the graph of y = xⁿ/(x²+1) have two points of zero gradient?
(I've already worked out the derivative which is [x^n-1((n-2)x²+n)]/(x²+1)²....as far as I got)

ii) State the coordinates of these points of zero gradient

thanks so much for all the help everyone!

humph · « **Reply #13 on:** December 31, 2007, 02:52:23 pm »

for the first part, you have to show for what values of $n$ that $f'(x) = 0$ has at least two distinct solutions.

$f(x)=\frac{x^n}{\left(x^2 + 1\right)}$
$\Rightarrow f'(x) =\frac{x^{n-1}((n-2)x^2+n)}{\left(x^2 + 1\right)^2}$

to find the points of zero gradient, we solve

$f'(x)=0$
$\Rightarrow f'(x) =\frac{x^{n-1}((n-2)x^2+n)}{\left(x^2 + 1\right)^2}=0$

the denominator is never zero, so the gradient is continuous and we can just multiply through.

$\Rightarrow x^{n-1}((n-2)x^2+n)=0$

$f$ has two points of zero gradient if the above equation has two distinct solutions.

if $n=1$ , then the equation above simplifies out to

$-x^2+1=0$

which clearly has the two solutions $x=1$ and $x=-1$ .

now suppose $n=2$ . then the equation simplifies out to

$2x=0$

this has only the one solution $x=0$ .

now suppose $n>2$ . then clearly $x=0$ is a solution to the equation. however, if

$(n-2)x^2+n=0$

$\Rightarrow x^2=-\frac{n}{n-2}$

which is obviously negative, as $n>2$ , so this has no real solutions

so if $n>1$ , the only point of zero gradient is $x=0$

if $n=1$ , then there are two points of zero gradient: $x=1$ and $x=-1$ .

to find the coordinates of these points of zero gradient, we simply substitute these two values back into $f$ .

$f(1)=1/2$

$f(-1)=-1/2$

so the coordinates of the points of zero gradient are $(1,1/2)$ and $(-1,-1/2)$

bec · « **Reply #14 on:** December 31, 2007, 03:50:39 pm »

thanks! ahhh i love this site. if i could use the karma thing you'd all be +ing all over the place.

just wondering, how do you use the maths font where you can write fractions and pi etc?

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