Question

1. Where is the point of intersection of the graphs of e⁻ᵗ and eᵗ⁺¹?

A. t= 1/2
​B. t=1
C. t=−1
D. t= -1/2
​
2. Graph the functions f(x)=ln(eˣ) and g(x)=eˡⁿ⁽ˣ⁾ and interpret the results.
A. The graphs are concave up and concave down parabolas.
B. The graph of f(x) is a straight line parallel to the x axis and the graph of g(x) is a straight line perpendicular to the y axis.
C. The graphs are straight lines intersecting at a right angle.
D. The graphs are straight lines identical to the graph of the equation y=x.

Answer 1

The point of intersection for the functions e⁻ᵗ and eᵗ⁺¹ is at t = -1/2. The functions f(x)=ln(eˣ) and g(x)=eᴉ⁻ᵒ(x) simplify to y=x, resulting in graphs that are identical straight lines.

Step-by-step explanation:

To find the point of intersection of the graphs of e⁻ᵗ and eᵗ⁺¹, we set the two functions equal to each other:

e⁻ᵗ

=

eᵗ⁺¹

Taking the natural logarithm (ln) of both sides to solve for t:

-ln(t) = t+1

This simplifies to:

-1 = 2t

The solution is t = -1/2, which corresponds to option D. For the functions f(x)=ln(eˣ) and g(x)=eᴉ⁻ᵒ(x), these are inverse functions since the natural logarithm and the exponential function are inverses of each other.

The function f(x)=ln(eˣ) simplifies to f(x)=x, and g(x)=eᴉ⁻ᵒ(x) simplifies to g(x)=x as well. Therefore, both graphs are straight lines that are identical to the graph of the equation y=x, matching option D.