Question text Find the inverse function of the exponential function [math: f(x)=2e5x−2+1]\displaystyle f(x)={2\,e^{5\,x-2}+1}. Note: Type [math: ln(a)] to enter [math: loge(a)]log_{e}(a). [math: f−1(x)=]f^{-1}(x)=[input] Check Question 7

Question

BlackTom AI · Accepted Answer

We are given the function f(x) = 2 e^{5x - 2} + 1 and asked to find its inverse.

First, set y = f(x): y = 2 e^{5x - 2} + 1.
Subtract 1 from both sides: y - 1 = 2 e^{5x - 2}.
Divide by 2: (y - 1)/2 = e^{5x - 2}.
Take the natural logarithm of both sides: ln((y - 1)/2) = 5x - 2.
Solve for x by adding 2 to both sides and then dividing by 5: x = [ln((y - 1)/2) + 2] / 5.

To express the inverse function f^{-1} in terms of x (replacing y with x): f^{-1}(x) = [ln((x - 1)/2) + 2] / 5.

Now, consider potential pitfalls:
- Forgetting to subtract 1 or 2 at the correct step would lead to an incorrect expression, such as missing the +2 inside or outside the logarithm.
- Misplacing the division by 5 or applying the logarithm to the wrong expression would yield a different form.
- The domain consistency: since e^{5x - 2} > 0, we have (y - 1)/2 > 0, so y > 1. Therefore the inverse is defined for x > 1 in the original function's range, which corresponds to the domain of f^{-1}.
- The presence of the natural log is appropriate here because the exponential was base e; using log base 10 would be incorrect in this context.

In this problem, the provided expression (ln((x - 1)/2) + 2) / 5 exactly matches the derived inverse, so it is the correct form for f^{-1}(x).

Question text Find the inverse function of the exponential function [math: f(x)=2e5x−2+1]\displaystyle f(x)={2\,e^{5\,x-2}+1}. Note: Type [math: ln(a)] to enter [math: loge(a)]log_{e}(a). [math: f−1(x)=]f^{-1}(x)=[input] Check Question 7多项填空题

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