Cal help

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warsatan

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Sep 12, 2005
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got this problem, need some help, thanks :)

Specify all lines through the point (1,5) and tangent to the curve f(x) = 3x^3+ x + 4.

many thanks :)
 
Specify all lines through the point (1,5) and tangent to the curve
f(x) = 3x^3+ x + 4
try this for a start


f'(x) = 9x^2 + 1

sub in x= 1

f'(x) = 10

ie the slope of the line is 10

y=mx + b where m = 10 and using (1,5)

5= 10 (1) + b
b= - 5

then
y= 10x - 5
 
can you take another derivative (sp) of f'(x) = 9x^2 + 1 and find another equation off this? just wonder.
 
apm said:
Specify all lines through the point (1,5) and tangent to the curve
f(x) = 3x^3+ x + 4
try this for a start


f'(x) = 9x^2 + 1

sub in x= 1

f'(x) = 10

ie the slope of the line is 10

y=mx + b where m = 10 and using (1,5)

5= 10 (1) + b
b= - 5

then
y= 10x - 5
Click to expand...

y=10x-5 is not the line that tangent to f(x), i just graphed it, any idea?
 
If you run x=1 through the original equation it should tell you if (1,5) actually lies on the line....
f(x)=3x^3 + x + 4
f(x)=3 + 1 + 4 = 8, so no, point 1,5 doesn't lie on the line, so the equation probably will be wrong.
 
Suppose that (a,b) is the point on the graph of f(x) at which a line through (1,5) is tangent.
Because f’(a)=9a<SUP>2</SUP>+1 that must be the slope.
Now solve (b-5)/(a-1) = 9a<SUP>2</SUP>+1 and b=3a<SUP>3</SUP>+a+4.
That will give you the point (a,b).
 
could you please give some more detail?
you said f’(a)=9a2+1 that must be the slope.

shouldnt you subs. (1,5) to get the slope?
 
That is easy. Note that (a,b) is on the graph of f(x).
The slope of the tangent line at (a,b) is the value of the derivative: f’(a).
You want a line through (1,5) which is tangent to the graph of f(x).

The slope of the line through (a,b) and (1,5) is (b-5)/(a-1).
 
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