If f(x)=(3x)/(1+x2) find f′(4)______.
Use this to find the equation of the tangent line to the curve y=(3x)/(1+x2) at the point (4,0.70588). The equation of this tangent line can be written in the form y=mx+b where m is: ________ and where b is:______
I will assume you meant
f(x) = 3x/(1 + x^2)
then by the quotient rule
f'(x) = ( 3(1 + x^2) - 3x(2x) )/(1 + x^2)^2
f'(4) = (3(17) - 12(8))/289 = -45/289
so now you have the slope m = -45/289
and your strange point (4, .70588)
Use your grade 9 method of y=mx + b
to find the equation.
To find the derivative of the function f(x) = (3x)/(1+x^2), we can use the quotient rule.
The quotient rule states that if we have a function in the form f(x) = g(x)/h(x), then the derivative of f(x) is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x))/[h(x)]^2
Applying this rule to our function f(x) = (3x)/(1+x^2), we have:
g(x) = 3x and h(x) = 1 + x^2
g'(x) = 3 and h'(x) = 2x
Substituting these into the quotient rule, we get:
f'(x) = [(3)(1 + x^2) - (3x)(2x)] / [(1 + x^2)^2]
Simplifying further, we have:
f'(x) = [3 + 3x^2 - 6x^2] / [(1 + x^2)^2]
f'(x) = (-3x^2 + 3) / [(1 + x^2)^2]
Now, to find f'(4), we substitute x = 4 into our expression for f'(x):
f'(4) = [-(3(4)^2) + 3] / [(1 + (4)^2)^2]
f'(4) = [-48 + 3] / [(1 + 16)^2]
f'(4) = [-45] / [17^2]
f'(4) = -45 / 289
Therefore, f'(4) = -45/289.
To find the equation of the tangent line to the curve y = (3x)/(1 + x^2) at the point (4, 0.70588), we can use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point and m is the slope of the tangent line.
We already know the slope, m = -45/289 from the derivative calculation.
Substituting the coordinates of the given point (4, 0.70588) into the point-slope form, we have:
y - 0.70588 = (-45/289)(x - 4)
Simplifying, we have:
y - 0.70588 = (-45/289)x + (45/72)
y - 0.70588 = (-45/289)x + 5/8
y = (-45/289)x + 5/8 + 0.70588
y = (-45/289)x + 46/58 + 0.70588
y = (-45/289)x + (46 + 805)/58
Finally, we can write the equation of the tangent line as:
y = (-45/289)x + 851/58
Therefore, the equation of the tangent line can be written in the form y = mx + b as:
m = -45/289
b = 851/58