Find the original functions f(x) and g(x) at x = 3 given the line tangents of both of those functions.

Question

If the graph of f(x) at x = 3 has a line tangent y = 4x + 2 and the graph of g(x) at x = 3 has a line tangent y = 5x - 4, what is f(x)? What is g(x)? What is (a) below? What is (b) below? What are the line tangents to the curves below at x = 3?

a) y = f(x) g(x) Answer: y = 114x – 188. 

b) y = f(x) / g(x) Answer:  y = -(26/121)x + 232/121.

Thanks, AL

Answer 1

f(x) has a slope at =3 of y'=4 g(x) has a slope at x=3 of y'=5

An infinite number of functions could satisfy those slopes at x=3, including the linear given tangent lines, f(x)=y=4x+2 and g(x)=y=5x-2 An infinite number of quadratic functions also could satisfy those slopes at x+3.

Assuming you want the simplest linear functions, then part a would be

f(x)g(x)= (4x+2)(5x-4)=20x²-6x-8

take the first derivative to get the slope: 40x-6 plug in x=3 to get 120-6=114

y=114x + y-intercept

y intercept = y(0)=-8

so tangent line at x=3 is y=114x-8

Part b:

f(x)/g(x)=(4x+2)/(5x-4)=(4x+2)(5x-4)^-1

take the first derivative, 1st term times derivative of 2nd term + 2nd term times derivative of 1st term

(4x+2)(-1)(5x-4)^-2(5) + (5x-4)^-1(4) = slope

plug in x=3 to find the slope of the tangent line:

14(-5)(11)^-2 +4/11 = -70/121 + 44/121 = -26/121

find the y-intercept, set x=0

y(0)=2(-5)(-4)^-2 + 4/-4 = -10/16 -1 = -26/16= -13/8

tangent line is y=-26x/121 - 13/8

but there are infinitely many other functions, quadratic, cubic and higher powered f(x) and g(x) functions

that also have the tangent lines in the given problem