Find the equation of the line tangent to the function h(x) = (ln(x) + 3)/((x^3) + 1) at point x=1.?

Question

thanks!

Answer 1

The formula for the equation to the tangent is f(a) + f '(a)(x-a)
and (a) is just the point at where you are trying to find the equation to the tangent, so it is 1(Because x=1). So...
You already have (a) for the formula so find f(a) and f '(a):

f(a) =>f(1)=(ln(x)+3)/((x^3)+1)=>
(0+3)/(1+1) =>3/2

f '(a) =>f '(1)=
(Take the derivative of the function then plug in 1 for x) =>

(1/x)(x^3 +1) - (ln(x)+3)(3x^2) / [(x^3) +1]^2 =>
(1/1)(1^3 +1) - (ln(1)+3)(3(1)^2) / [(1^3) +1]^2 =>
2 - ( 0 +3) (3) / 2^2=>
(2-9)/4 => -7/4

Now plug that in to the formula that I gave you at the top line
and you have:
(3/2) +(-7/4)(x-1)

Answer 2

To find the equation of a tangent line to a curve (y = h(x)) at a given point on the curve we proceed as follows
Find the slope of the tangent at that point which is given by dy/dx = dh(x)/dx at the given point on the curve y = hf(x)and then using the point and the slope find the equation of the tangent.
In the above case the x coordinate is given then we find y coordinate by subsituting x =1
y = h(x) = (ln(x) + 3) / (x^3 +1)
h(1) = (ln (1) +3)/(1^3+1) = 3/2 because ln 1 = log1 to the base e
= 0 and 1^3 = 1
hence the point where we find dy/dx is the point (1,3/2) which is the point on the curve and the line and is also the point where the tangent touches the curve
Now we find d h(x)/dx = [(x^3) + 1)d(ln(x) + 3)/dx - (ln(x) + 3))d((x^3)+1)/dx ] /(x^3)+1)^2 (using the rule of finding derivative for divisions of two functions u/v)
at point (1,3/2) the value of derivative or slope d h(x)/dx is
[(x^3) + 1)(1/x) - (ln(x) + 3)(3x^2)]/(x^3) + 1)^2
= [(1^3) + 1)(1/x) - (ln(1) + 3))(3(1^2)]/(1^3) + 1)^2
= [ (2)(1/1) - (0 +3)(3)]/[1+1)^2 = [2-9]/2*2 = (-7/4)
with slope -7/4 and point (1,3/2) the equation is given by
Y-3/2 =(-7/4)(X-1)
or 4Y - 6 = -7X +7
7X + 4Y -13 = 0
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we have used following formulae
d[u(x)/v(x)]/dx = (v du/dx - udv/dx)/v^2
and
d[ln(x)]/dx = 1/x

Answer 3

h=ln+3/(x^1)+x

Answer 4

slope = ((x^3 + 1)(1/x)-(ln(x)+3)(3x^2))/((x^3)+1)^2

at x = 1 slope = -1.75

equation of line at x= 1

y= -1.75(x -1)