If x=e^t and y=t^2-3, then for what values of t, t>=0, is the function y=f(x) concave up?
let's eliminate the parameter t
from x = e^t
lnx = t
then y = (lnx)^2 - 3
to be concave up, the second derivative must be positive
y' = 2(lnx)(1/x)
y'' = 2lnx (-1/x^2) + (1/x)(1/x)
= (-2lnx + 1)/x^2
x^2 is always positive
so -2lnx + 1 > 0
lnx < -1/2 ----> x < .6065...
so e^t < .6065
t < ln .6065
t < -.5
x=e^t and y=t^2-3
dx/dt = e^t
dy/dt = 2t
dy/dx = (dy/dt)/(dx/dt) = 2t/e^t
d^y/dx^2 = d/dt (dy/dx) * dt/dx
d/dt(dy/dx) = -2(t-1)/e^t
dt/dx = 1/(dx/dt) = 1/e^t
d^2y/dx^2 = -2(t-1)/e^2t
e^2t is always positive, so y">0 when t<1
Hmmm. From above,
y' = 2lnx/x
y" = 2lnx (-1/x^2) + 2(1/x)(1/x)
= 2(1-lnx)/x^2
so, y" > 0 when lnx < 1
or, when t < 1
To determine when the function y=f(x) is concave up, we need to analyze the second derivative of y with respect to x. But first, we need to find the derivative of y with respect to x.
Given that x = e^t, we can write t = ln(x). Differentiating this equation with respect to x, we get:
dt/dx = 1/x.
Now, let's differentiate y = t^2 - 3 with respect to x using the chain rule:
dy/dx = (dy/dt) * (dt/dx).
To find dy/dt, we differentiate y = t^2 - 3 with respect to t:
dy/dt = 2t.
Substituting dt/dx = 1/x and dy/dt = 2t back into dy/dx, we get:
dy/dx = (2t) * (1/x) = 2t/x.
To find the second derivative, we differentiate dy/dx with respect to x:
d^2y/dx^2 = d/dx (2t/x).
Using the quotient rule, we differentiate t with respect to x:
(dt/dx) = 1/x.
Differentiating 2t with respect to x gives:
d(2t)/dx = 2 * (dt/dx) = 2 * (1/x) = 2/x.
Finally, differentiating x with respect to x gives:
(dx/dx) = 1.
Applying the quotient rule, we get:
d^2y/dx^2 = (2/x - (2/x^2) * t) * 1 - (2t/x) * 1.
Simplifying this expression, we have:
d^2y/dx^2 = 2/x - 2t/x^2 - 2t/x.
We want to determine the values of t (t >= 0) for which the second derivative is positive, indicating that the function y = f(x) is concave up.
Setting d^2y/dx^2 > 0, we get:
2/x - 2t/x^2 - 2t/x > 0.
Combining the terms, we have:
2/x - (2t/x^2) - (2t/x) > 0.
To isolate t, we can multiply the entire inequality by x^2:
2(x^2/x) - (2t) - 2t(x) > 0.
Simplifying, we get:
2x - 4t - 2tx > 0.
Rearranging the terms, we have:
2x - 2tx > 4t.
Factoring t out, we get:
t(2 - 2x) > 4t.
Since t >= 0, we can divide both sides of the inequality by t:
2 - 2x > 4.
Simplifying, we have:
-2x > 2.
Dividing the inequality by -2 (and reversing the inequality sign), we get:
x < -1.
Since x = e^t, this inequality implies that t < ln(-1). However, the natural logarithm is only defined for positive numbers, so ln(-1) is undefined.
Therefore, there are no values of t (t >= 0) for which the function y = f(x) is concave up.