SOLUTION: Find the value of x for which the tangent line to {{{y=2x^2 + 1}}} will be parallel to the tangent line to {{{y = 4ln(x) - 3}}}

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Question 1153638: Find the value of x for which the tangent line to y=2x%5E2+%2B+1 will be parallel to the tangent line to y+=+4ln%28x%29+-+3
Found 2 solutions by greenestamps, jim_thompson5910:
Answer by greenestamps(13195) About Me  (Show Source):
You can put this solution on YOUR website!


y+=+2x%5E2%2B1
dy%2Fdx+=+4x

y+=+4ln%28x%29%2B3
dy%2Fdx+=+4%2Fx

The slopes are equal when
4x+=+4%2Fx
x%5E2=1
x+=+1 or x+=+-1

The second function is undefined at x=-1; so x=1 is what we are looking for.

For the first function,
f%281%29+=+2%281%5E2%29%2B1+=+2%2B1+=+3

For the second function,
f%281%29+=+4ln%281%29%2B3+=+4%280%29%2B3+=+3

The slopes of the two graphs are the same at (1,3); and the two graphs are tangent to each other at that point.

At x=1, the slope of both graphs is 4; the line with slope 4 passing through (1,3) is y+=+4x-1

A graph, showing the two curves and the common tangent line:

graph%28400%2C400%2C-1%2C2%2C-1%2C6%2C2x%5E2%2B1%2C4ln%28x%29%2B3%2C4x-1%29

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

The tutor @greenestamps has a great set of steps and the correct final answer. The only issue I see is that somehow 4ln(x)-3 turned into 4ln(x)+3. This will mean the two functions do not intersect at (1,3). Other than that, everything else is perfect.