Calculate the approximate value of (1.01)(1.05)^2 e^1.1+(1.01)ln(1.05) using the verification of the three-variable function

Let's denote the three variables as follows:

x = 1.01 , y = 1.05 , z = 1.1

The given expression can be rewritten as:

f(x, y, z) = x y^2 e^z + x \ln(y)

Substitute the given values of x , y , and z into the function f(x, y, z) :

f(1.01, 1.05, 1.1) = 1.01 \times 1.05^2 \times e^{1.1} + 1.01 \times \ln(1.05)

Now, calculate the values:

f(1.01, 1.05, 1.1) = 1.01 \times 1.1025 \times 3.004166 + 1.01 \times \ln(1.05)

f(1.01, 1.05, 1.1) = 1.1426 + 1.01 \times \ln(1.05)

Now, use the third variable function to simplify:

Let's define another function as g(t) = 1.01 \times \ln(t)

So, g(1.05) = 1.01 \times \ln(1.05)

Therefore, the final function becomes:

f(1.01, 1.05, 1.1) = 1.1426 + g(1.05)

Now calculate the value of g(1.05) :

g(1.05) = 1.01 \times \ln(1.05) \approx 1.01 \times 0.04879 \approx 0.0492769

Finally, substitute this value into the original function:

f(1.01, 1.05, 1.1) = 1.1426 + 0.0492769 \approx \boxed{1.19188}