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Question: Let f: R$^+$ $\rightarrow$ R defined as f(x) = e$^x$ + lnx and g = f$^{-1}$ then correct statement(s...

Let f: R+^+ →\rightarrow R defined as f(x) = ex^x + lnx and g = f−1^{-1} then correct statement(s) is/are-

A

g''(e) = 1−e(1+e)3\frac{1-e}{(1+e)^3}

B

g''(e) = e−1(1+e)3\frac{e-1}{(1+e)^3}

C

g'(e) = e+1

D

g'(e) = 1e+1\frac{1}{e+1}

Answer

The correct statements are (A) and (D).

Explanation

Solution

Let the given function be f(x)=ex+ln⁡xf(x) = e^x + \ln x defined for x∈R+=(0,∞)x \in R^+ = (0, \infty). The inverse function is g=f−1g = f^{-1}. We need to evaluate g′(e)g'(e) and g′′(e)g''(e).

First, find the value of xx such that f(x)=ef(x) = e. ex+ln⁡x=ee^x + \ln x = e. By inspection, for x=1x=1, we have e1+ln⁡1=e+0=ee^1 + \ln 1 = e + 0 = e. So, f(1)=ef(1) = e. This means g(e)=1g(e) = 1.

Next, find the first derivative of f(x)f(x): f′(x)=ddx(ex+ln⁡x)=ex+1xf'(x) = \frac{d}{dx}(e^x + \ln x) = e^x + \frac{1}{x}. Evaluate f′(x)f'(x) at x=1x=1: f′(1)=e1+11=e+1f'(1) = e^1 + \frac{1}{1} = e+1.

The formula for the derivative of an inverse function g(y)=f−1(y)g(y) = f^{-1}(y) is g′(y)=1f′(x)g'(y) = \frac{1}{f'(x)}, where y=f(x)y = f(x). We want to find g′(e)g'(e). Here y=ey=e, which corresponds to x=1x=1. g′(e)=1f′(1)=1e+1g'(e) = \frac{1}{f'(1)} = \frac{1}{e+1}.

Now, find the second derivative of f(x)f(x): f′′(x)=ddx(ex+x−1)=ex−x−2=ex−1x2f''(x) = \frac{d}{dx}(e^x + x^{-1}) = e^x - x^{-2} = e^x - \frac{1}{x^2}. Evaluate f′′(x)f''(x) at x=1x=1: f′′(1)=e1−112=e−1f''(1) = e^1 - \frac{1}{1^2} = e - 1.

The formula for the second derivative of an inverse function is g′′(y)=−f′′(x)(f′(x))3g''(y) = -\frac{f''(x)}{(f'(x))^3}, where y=f(x)y = f(x). We want to find g′′(e)g''(e). Here y=ey=e, which corresponds to x=1x=1. g′′(e)=−f′′(1)(f′(1))3=−e−1(e+1)3=−(e−1)(e+1)3=1−e(e+1)3g''(e) = -\frac{f''(1)}{(f'(1))^3} = -\frac{e-1}{(e+1)^3} = \frac{-(e-1)}{(e+1)^3} = \frac{1-e}{(e+1)^3}.

Comparing our results with the given options: (A) g′′(e)=1−e(1+e)3g''(e) = \frac{1-e}{(1+e)^3}. This matches our result. (B) g′′(e)=e−1(1+e)3g''(e) = \frac{e-1}{(1+e)^3}. This does not match our result. (C) g′(e)=e+1g'(e) = e+1. This does not match our result. (D) g′(e)=1e+1g'(e) = \frac{1}{e+1}. This matches our result.

Therefore, the correct statements are (A) and (D).

Explanation:

  1. Find xx such that f(x)=ef(x) = e. ex+ln⁡x=e  ⟹  x=1e^x + \ln x = e \implies x=1. Thus g(e)=1g(e)=1.
  2. Calculate f′(x)=ex+1/xf'(x) = e^x + 1/x and f′′(x)=ex−1/x2f''(x) = e^x - 1/x^2.
  3. Evaluate f′(1)=e+1f'(1) = e+1 and f′′(1)=e−1f''(1) = e-1.
  4. Use the formula g′(y)=1/f′(x)g'(y) = 1/f'(x) at y=e,x=1y=e, x=1. g′(e)=1/f′(1)=1/(e+1)g'(e) = 1/f'(1) = 1/(e+1).
  5. Use the formula g′′(y)=−f′′(x)/(f′(x))3g''(y) = -f''(x)/(f'(x))^3 at y=e,x=1y=e, x=1. g′′(e)=−f′′(1)/(f′(1))3=−(e−1)/(e+1)3=(1−e)/(e+1)3g''(e) = -f''(1)/(f'(1))^3 = -(e-1)/(e+1)^3 = (1-e)/(e+1)^3.
  6. Match the results with the given options.