2^2y - 2×2^ye^x + e^2x

= 4^y - 2^y+1e^x + e^2x

6 × 3^x2^x = 6×(3×2)^x

= 6^x+1

f(2) - f(0) =

(2e² + 3e⁴ - 1) - (2 + 3 - 1)

=2e² + 3e⁴ - 5

Both growth curves:

• pass through (0, 1);
• y = 3^x climbs less slowly and is below y = 5^x for x > 0;
• y = 3^x falls slower and is therefore above y = 5^x for x < 0.

The basic exponential growth curve
(y = e^x) rises as x increases.

The transformation of "+ 3" moves the curve up so that now it has y = 3 as its asymptote rather than the x-axis.

The y intercept is at y = e ⁰ + 3 = 4.
At x = 1, y = 3 + e = 3 + 2.7 = 5.7.

The negative sign in front of the exponential term turns the graph of the decay curve upside down.

The addition transformation of +3 moves the curve up 3 units so the asymptote rises from y = 0 to y = 3.

The coefficient and index both of 2 make the curve fall more rapidly.

The basic exponential growth curve
(y = -e^-x) rises as x increases.

The addition transformation of +1 moves the curve up 1 unit so the asymptote rises from y = 0 to y = 1.

The coefficient of 2 makes the curve rise more quickly as x increases.

The catenary adds both exponential functions together. Hence the y intercept equals (1 + 1)÷ 2 = 1.

The addition also means that the curve on both sides of the y axis has a larger y value than each of the two basic exponential curves.

Clearly also the function is even.

At the intersection of the two curves, the y values are the same. Hence we replace the exponential function in the combined equation and put the equation - 0 as there is no difference between the y values.

f(x) = e^-x - x + 1 = y - x + 1 = 0

So we obtain y = x - 1

As there is only one point of intersection, the combined equations have exactly one root.