Answer:
a) y = 7.74*x + 7.5
b) y = 1.148*x + 6.036
Step-by-step explanation:
Given:
f(x) = 6 - 10*x^2
g(x) = 8 - (x-2)^2
Find:
(a) The equation of the line with the LARGEST slope that is tangent to both graphs is
(b) The equation of the second line tangent to both graphs is:
Solution:
- First, let's calculate the derivatives for the two functions provided:
f'(x) = -20*x
g'(x) = -2*(x-2)
- Given that the derivatives of both functions are dependent on the x-value, we will pick a common point x_o for both f(x) and g(x). This point is ( x_o, g(x_o)). Thus,
g'(x_o) = -2*(x_o - 2)
- Next, we will determine the slope of a line that is tangent to both graphs at point (x_o, g(x_o) ) on g(x) and at ( x, f(x) ) on f(x):
m = (g(x_o) - f(x)) / (x_o - x)
m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)
m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)
m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)
m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)
- Now we need to set the slope from our equation equal to the derivatives we calculated earlier for each function:
m = f'(x) = g'(x_o)
- We will work through the first expression:
m = f'(x)
( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x
Eq 1. (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2
And,
m = g'(x_o)
( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x
-2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)
Eq 2 -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)
- Now we can subtract the two equations (Eq 1 - Eq 2):
-20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0
-22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0
- Rearranging gives us: 20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0
20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0
(x - x_o)(20*x - 2*x_o + 4) = 0
x = x_o, x_o = 10x + 2
- For x_o = 10x + 2,
(g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x
(8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x
(-90*x^2 + 2) = -180*x^2 - 40*x
90*x^2 + 40*x + 2 = 0
- Now solving the above quadratic equation:
x = -0.0574, -0.387
- The maximum slope occurs at x = -0.387, with the line’s equation being:
y - 4.502 = -20*(-0.387)*(x + 0.387)
y = 7.74*x + 7.5
- The second tangent line is:
y - 5.97 = 1.148*(x + 0.0574)
y = 1.148*x + 6.036
