Values of x - By differentiating :D (1 Viewer)

[Smile12345]

Hello All...

Could someone please help me with this question...

Q. For what value of x does the functions have y = 1/4x -1 have y'= -4/49

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(Sorry I can't work Latex - This what I tried to do in Latex

 

[nerdasdasd]

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[nerdasdasd]

 

[Trebla]

Differentiate to find y' and set that to equal -4/49 and solve for x. You should get two solutions of x.

 

[Carrotsticks]

View attachment 28453

You should get two solutions of x.

.

 

[Parvee]

you should get x=2 and -3/2

 

[Smile12345]

So y'= -4(4x + 1)^-2.... This right?

 

[Smile12345]

you should get x=2 and -3/2

I got x = -2 and 3/2... from x= -1 + - root 49 / 4.... ==> x = -2, 3/2

This right or not?

 

[Smile12345]

Thanks all for your assistance. (and even two moderators!)

 

[Parvee]

So y'= -4(4x + 1)^-2.... This right?

y'=-4(4x-1)^-2
and you are given y'=-4/49
So you can equate them -4/49=-4/(4x-1)^2
Which gives you (4x-1)^2=49 and then you just solve the quadratic

 

[Smile12345]

y'=-4(4x-1)^-2
and you are given y'=-4/49
So you can equate them -4/49=-4/(4x-1)^2
Which gives you (4x-1)^2=49 and then you just solve the quadratic

Yes, I see. I had done +1 instead of -1, thanks for your continued help.

This is a different topic but....How do I do the below? I am sure there is a way that doesn't involve using a calculator. Am I right?

Simplify
a) tan 70(deg) + cot 20(deg) - 2tan70(deg)
b) cot25 + tan65 / cot25

Thanks.

 

[Parvee]

I got x = -2 and 3/2... from x= -1 + - root 49 / 4.... ==> x = -2, 3/2

This right or not?

uhm is your original equation y=1/(4x-1) or y=1(4x+1) :L

if its 1/(4x-1) you will get x= 2 and -3/2
if its 1/(4x+1) you will get x= -2 and 3/2

 

[Smile12345]

uhm is your original equation y=1/(4x-1) or y=1(4x+1) :L

if its 1/(4x-1) you will get x= 2 and -3/2
if its 1/(4x+1) you will get x= -2 and 3/2

Thanks again... Yes I just realised that... Sorry... And then I posted at the same time as you.

 

[Parvee]

Yes, I see. I had done +1 instead of -1, thanks for your continued help.

This is a different topic but....How do I do the below? I am sure there is a way that doesn't involve using a calculator. Am I right?

Simplify
a) tan 70(deg) + cot 20(deg) - 2tan70(deg)
b) cot25 + tan65 / cot25

Thanks.

for a) cot(theta)=tan(90-theta)
therefore cot(20)=tan(70)
So tan(70)+tan(70)-2tan(70)=0

b) tan65/cot25=1 using cot(theta)=tan(90-theta)
so you end up with cot(25)+1 (is this the right answer? :L)

 

[Smile12345]

for a) cot(theta)=tan(90-theta)
therefore cot(20)=tan(70)
So tan(70)+tan(70)-2tan(70)=0

b) tan65/cot25=1 using cot(theta)=tan(90-theta)
so you end up with cot(25)+1 (is this the right answer? :L)

Thanks... Unfortunately I don't have the answers... It's a question from a past paper. If I do it on the calculator I get two??? But not sure if this is right....

 

[Parvee]

Thanks... Unfortunately I don't have the answers... It's a question from a past paper. If I do it on the calculator I get two??? But not sure if this is right....

The answer for b) is correct, I was just asking because they maybe gave the answer in a different format

 

[Smile12345]

Parvee (if you're still there)...

I think I have got it...

cot25 = tan65 (same reason as what you gave - Cot(theta) = tan(90 - theta)

So tan 65 + tan65/ cot 25

Then we can replace the bottom with tan65

So 2tan65/ tan65 = 2 (because the tan65) cancel out!

 

[Smile12345]

Yeah ok then. Thanks for the consideration... Unfortunately I don't have the answers. :L

 

[Parvee]

Parvee (if you're still there)...

I think I have got it...

cot25 = tan65 (same reason as what you gave - Cot(theta) = tan(90 - theta)

So tan 65 + tan65/ cot 25

Then we can replace the bottom with tan65

So 2tan65/ tan65 = 2 (because the tan65) cancel out!

hmm not quite right
doing so you would get tan(65)+(tan(65)/tan(65))
which give tan(65)+1

 

[Smile12345]

hmm not quite right
doing so you would get tan(65)+(tan(65)/tan(65))
which give tan(65)+1

oh, i'm not sure. I thought that if I just replaced both cot25 with tan65 (for the reason we both know they're the same) so it would be tan65 + tan65 / tan 65.