Find k such that the line y = 2x + 18 is tangent to the graph of the function below.
y=k√x
k=?

- amy0799

Find k such that the line y = 2x + 18 is tangent to the graph of the function below.
y=k√x
k=?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- Loser66

I told you, take derivative of the curve \(y = k\sqrt x\), what do you get?

- amy0799

1/2k^-1/2

- Loser66

|dw:1440973881338:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- freckles

|dw:1440974118718:dw|
you want y=2x+8 and y=ksqrt(x) to have a common point (a,f(a))
and you also want the derivative of 2x+8 to be equal to the derivative of ksqrt(x) at (a,f(a))

- amy0799

right, I understand that, I just don't know how to find it.

- freckles

you have
2a+8=k sqrt(a) <--since we want 2a+8 to be equal to k sqrt(a)
and you also have
2=k/(2sqrt(a))

- Loser66

Thanks @freckles. Shame on me, I guided him wrong. @amy0799 I am sorry.

- freckles

it is just a system of equations to solve

- amy0799

\[\frac{ 2x+8 }{ \sqrt{x}} = k\]

- amy0799

what do I do next?

- freckles

\[2a+8=k \sqrt{a} \\ 2=\frac{k}{2 \sqrt{a}} \implies \sqrt{a}=\frac{k}{4} \implies a=\frac{k^2}{16} \text{ insert into first equation }\]

- freckles

doing that will give you an equation just in terms of k

- amy0799

insert it into which equation?

- freckles

the First equation

- freckles

the first equation I wrote above the second equation

- freckles

\[2a+8=k \sqrt{a} \\ 2=\frac{k}{2 \sqrt{a}} \implies \sqrt{a}=\frac{k}{4} \implies a=\frac{k^2}{16} \text{ insert into first equation } \]
there are equations I'm playing with here
the first equation which is on the first line
and the second equation which on the second line
the second line I used the second equation to express sqrt(a) in terms o k
and I also expressed a in terms of k using the second equation
we can now write the first equation which I put on the first line in terms of k

- amy0799

so I plug a=k^2/16 into 2a+8=ksqrt(a)?

- freckles

yes you also plug in k/4 for sqrt(a)

- amy0799

\[2(\frac{ k^{2} }{ 16 })+8=k \sqrt{k \sqrt{a}}\]
this doesn't make sense

- freckles

you are write that equation doesn't make sense
you should have gotten something else entirely different

- freckles

\[2a+8=k \sqrt{a} \\ \text{ we determined } \sqrt{a}=\frac{k}{4} \text{ then we squared both sides \to get } a=\frac{k^2}{16} \\ \text{ plug \in both of these results we obtained from the second equation } \\ \text{ into the first } \\ 2(\frac{k^2}{16})+8=k(\frac{k}{4})\]

- freckles

now solve the equation for k
(you will see you have to discard a value that you get for k when solving this equation because of the equation sqrt(a)=k/4 which tells us that sqrt(a) is positive and so k has to be positive )

- freckles

do you know how to solve the above quadratic

- amy0799

not really

- freckles

k*k=k^2
and 2/16=1/8
you have:
\[\frac{k^2}{8}+8=\frac{k^2}{4} \\ \text{ you can multiply both sides by 8 \to get } \\ k^2+64=2k^2 \]
can you solve from here?

- amy0799

k=8?

- freckles

\[64=k^2 \\ k=8 \text{ or } -8 \\ \text{ but we did mentioned that } k>0 \\ \text{ so yes } k=8\]

- amy0799

thank you! can u help me with another one?

- freckles

i can try yes

- freckles

but

- freckles

before we go on you need to understand this one

- freckles

like do you have any questions? do you understand how I solved the system of equations?

- amy0799

yea I understood that

- amy0799

Consider the following function.
f(t) = 2t^2 − 3
(a) Find the average rate of change of the function below over the interval [3, 3.1].
(b) Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval.
(at t = 3)
(at t = 3.1)

- freckles

\[\text{average rate of change of } f \text{ from }t=a \text{ to } t=b \\ \text{ is given by the formula } \frac{f(b)-f(a)}{b-a}\]

- freckles

\[\text{ instantaneous rate of change of} f \text{ at } x=c \text{ is given by } f'(c)\]

- amy0799

so what's the first step I need to do?

- freckles

you used the formula above to find the average rate of change from t=3 to t=3.1...

- freckles

replace b with 3.1 and a with 3

- amy0799

\[\frac{ f(3.1)-f(3) }{ 3.1-3 }\]

- freckles

yes now evaluate f(3.1) and f(3)
and do the difference on top and bottom

- freckles

and then the division of the top by the bottom

- amy0799

I don't understand what you mean

- freckles

f(t) is given as 2t^2-3
you are to find f(3.1) and f(3) by using f(t)=2t^2-3
replace t with 3.1 and get a result for f(3.1)
replace t with 3 and get a result for f(3)

- amy0799

f(3.1)=16.22
f(3)=15

- freckles

\[\frac{f(3.1)-f(3)}{3.1-3}=\frac{16.22-15}{3.1-3}\]
now find the difference for both top and bottom

- freckles

the difference of 16.22 and 15 is another way to say 16.22-15
if you are confused by the word difference

- amy0799

12.2

- amy0799

is that the average rate of change?

- freckles

yes that is the average rate of change of f(t)=2t^2-3 from t=3 to t=3.1

- amy0799

I got part b wrong

- freckles

did you find the f'(3) and f'(3.1)

- amy0799

yea isn't it 16.22 and 15?

- freckles

no

- freckles

f' means to find the derivative of f

- freckles

and you just plugged into original function

- freckles

did you not look earlier at how I said to find the instantaneous rate of change?

- freckles

I posted the following two comments:
\[\text{average rate of change of } f \text{ from }t=a \text{ to } t=b \\ \text{ is given by the formula } \frac{f(b)-f(a)}{b-a}\]
\[\text{ instantaneous rate of change of} f \text{ at } x=c \text{ is given by } f'(c)\]

- amy0799

I don't understand it

- freckles

you don't know how to find the derivative of 2t^2-3?

- freckles

do you know power rule and constant rule?
if not use the definition of derivative

- amy0799

4t.

- freckles

yes so what do you not understand?

- freckles

you have f'(t)=4t
and you are asked to evaluate f'(t) at the endpoints

- freckles

the endpoints are 3 and 3.1

- freckles

f'(3)=?
f'(3.1)=?

- amy0799

12 and 12.4

- amy0799

my very first question. I got k=8 wrong

- freckles

did you look at what we wrote and what you asked?

- freckles

you should see we changed 18 to 8

- freckles

you know the line being y=2x+18 instead of y=2x+8

- amy0799

is the answer not 8?

- freckles

the answer is k=8 if the question involved y=2x+8
but the equation was y=2x+18
please look at our work and your question
our answer was for if the line was y=2x+8 which I started over and over
but the equation was y=2x+18
so just redo the work for the equation being y=2x+18 instead of it being y=2x+8

- freckles

which I stated over and over in the answer*

- freckles

do you understand?

- freckles

the question was:
Find k such that the line y = 2x + 18 is tangent to the graph of the function below.
y=k√x
k=?
however our work was for:
Find k such that the line y = 2x + 8 is tangent to the graph of the function below.
y=k√x
k=?

- freckles

you should be able to do your question now if you actually understand the work we did to find k for this other question

- amy0799

can you help me with it? @freckles

- freckles

I thought you said you understand the process?

- freckles

\[\text{ solve the system of equations } \\ (2x+18)'|_{x=a}=(k \sqrt{x})'|_{x=a} \\ 2a+18=k \sqrt{a}\]

- freckles

it is the same process we did above

- freckles

except there is just a little number change

- amy0799

I understood how you solved the quadratic, but not how you got a

- freckles

\[2a+8=k \sqrt{a} \\ 2=\frac{k}{2 \sqrt{a}} \implies \sqrt{a}=\frac{k}{4} \implies a=\frac{k^2}{16} \text{ insert into first equation } \]
is this what you are talking about?
you don't know how I solved the second equation for sqrt(a)?

- freckles

\[2=\frac{k}{2 \sqrt{a}} \\ \text{ I just multiplied } 2 \sqrt{a} \text{ on both sides } 2(2 \sqrt{a})=k \\ 4 \sqrt{a}=k \\ \text{ and then divided both sides by } 4\]

- freckles

and the only different and this set of equations is 8 is supposed to be 18

- freckles

\[2a+18=k \sqrt{a} \\ \sqrt{a}=\frac{k}{4} \text{ squaring both sides gives } a=\frac{k^2}{16} \text{ replace the items in the first equation }\]

- amy0799

k=18?

- freckles

how did you get that?

- freckles

did you replace a with k^2/16
and sqrt(a) with k/4?

- freckles

\[2a+18=k \sqrt{a} \\ 2(\frac{k^2}{16})+18=k(\frac{k}{4})\]

- freckles

solve the equation for k

- amy0799

\[\frac{ 18k^{2} }{ 8 }+324=\frac{ 18k ^{2} }{ 4 }\]
is this right so far?

- freckles

why did you choose to multiply both sides by 18

- amy0799

that's how u did it before

- freckles

no I multiplied both sides by something to clear the fractions not to make the problem more fractionie (I know that isn't a word)
\[\frac{2}{16}=\frac{1}{8} \\ k \cdot k=k^2 \\ \text{ you have } \\ 2(\frac{k^2}{16})+18=k(\frac{k}{4}) \\ \text{ so this is equivlant to } \\ \frac{k^2}{8}+18=\frac{k^2}{4} \\ \text{ if you multiply both sides by 8 } \text{ you can clear the fractions }\]

- freckles

the denominators were same in previous previous problem
and I chose to multiply both sides by 8 then so it shouldn't change here

- freckles

unless you want to make the problem more complicated looking :p

- amy0799

k=12

- freckles

yes
\[k^2+8(18)=2k^2 \\ 8(18)=k^2 \\ 16(9)=k^2 \\ 4(3)=k \\ k=12\]
and again we chose 12 instead of -12 because we wanted k to be positive
@amy0799 you might want to go and review how to solve algebraic equations
practice, practice and you can get better at this I promise

- amy0799

yeaa I know >.<
I have one more I need help with

- freckles

you can post it
but brb

- freckles

ok i'm here

- amy0799

Consider the following function.
\[f(x)=\sqrt{9-x^{2}}\]
Find the derivative from the left at x = 3. If it does not exist, enter NONE.
Find the derivative from the right at x = 3. If it does not exist, enter NONE.

- freckles

do you know what the function looks like?

- freckles

|dw:1440979836143:dw|
does the function exist to the right of x=3?

- amy0799

no

- freckles

so we already know one answer is does not exist...
now let's look at the left side:
\[\text{ \left derivative of } f \text{ at } x=3 \text{ is given by } \lim_{x \rightarrow 3^-}\frac{f(x)-f(3)}{x-3} \\ \]

- amy0799

it doesn't exist either

- freckles

correct

- amy0799

thank you so so much and I'm really sorry if I took up ur time!!

- freckles

It is only fine if you take up my time if you have learned something

- freckles

do you feel like you learned anything

- amy0799

I learned a lot. thank you!

- freckles

np :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.