MAV 2010 exam 1 help

[ldee]

Hey guys.

I don't understand the attached question. Usually they give a point to find the tangent at, but in this case they actually give the equation. I looked at the worked solutions but I'm still a bit sketchy, could somebody please go through it?

Thanks

[spongebob-7]

may be wrong but ill give it a shot,
g(x)= x-loge(x)
g'(x)=1-1/x
now we know the gradient is -1/2, so we need to find x that lets 1-1/x =-1/2
-1/x=-3/2
x=2/3
Now we sub that back into g(x) to get the y value
we get 2/3+loge(3/2)
now using the tangent formula y-y1=m(x-x1)
we get y-2/3-loge(3/2)=-1/2(x-2/3)
rearrange to get y by itself
y=- 1/2x +1+loge(3/2)
hence k= 1 + loge(3/2)

[FlorianK]

Gradient of tangent is -0.5
g'(x)=1-1/x
-0.5=1-1/x
1/x=3/2
x=2/3
g(2/3)=2/3-ln(2/3)
x=2/3 , y=2/3 - ln(2/3)=-(2/3)/2 +k
--> k=1-ln(2/3)

[BigAl]

you know the gradient. Using that you can find the point the original graph passes through. using all given information you can find the value of k as solved above.

[ldee]

Thanks a lot everyone, makes a lot of sense