posted by C .
Determine whether or not each of the following functions is invertible. Give your reasons for believing the function is invertible or not.
Please check this for me. I am not sure if I am adequately explaining my answer and if my answer is right.
a) y= log10(1 + 1/x)
y'= ((1/(1 + 1/x)*ln(10)) * (-1/x^2))
I plugged 100 and -100 into the derivative and got -4.3 X 10^-5 and - 4.4 X 10^-5
NOT INVERTIBLE because I plugged in -0.5 and got a positive answer, so the derivative is then increasing and decreasing, right? The function also has two inflection points at x=0 and x=-1
b) y= e^(x^2 - 5x + 6)
y'= (e^(x^2 - 5x + 6) * (2x - 5))
I plugged 10 and -10 into the derivative and got 3.1 X 10^25 and -1.4 X 10^69
The function is increasing and decreasing, so there is a max and min value and there is more than one x-value for the y-value, thus NOT INVERTIBLE.
Let's start with b, the answer of which is correct: it is not invertible.
Domain of the function is ℝ since it is the exponential function raised to a polynomial power.
If it were invertible, as you mentioned, the derivative would not change sign.
Change in sign of the derivative implies that on each side of the extremum, the function will take on the same value within its domain, which renders the function not invertible. In other words, it does not pass the horizontal line test.
For part a.
Again, we have to first determine the domain of the function:
Since log functions cannot take on negative values, we determine that the domain of the function is limited to the range of x where the expression inside the log is non-negative, namely (-&infin,-1)∪(0,∞).
Now we have to look at the function within its domain.
We find that the function is monotonically decreasing throughout its domain, within which it satisfies the horizontal line test.
Would you therefore re-evaluate your answer?
To help make your decision, you can have some thoughts on two things:
1. look at the graph of the function:
2. Find its inverse:
which is a perfectly legitimate function undefined at x=0.