Question
Approximate the function at the value of $x$ to four decimal places.(a) $f(x)=\ln (x+1)+e^{x}, \quad x=2$(b) $g(x)=\frac{(\log x)^{2}-\log x}{4}, \quad x=3.97$
Step 1
We substitute $x=2$ into the function to get $f(2)=\ln (2+1)+e^{2}$. Show more…
Show all steps
Your feedback will help us improve your experience
Priyanka Sadarangani and 89 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Approximate the function at the value of $x$ to four decimal places. (a) $f(x)=\log \left(2 x^{2}+1\right)-10^{-x}, \quad x=1.95$ (b) $g(x)=\frac{x-3.4}{\ln x+4}, \quad x=0.55$
Inverse, Exponential, and Logarithmic Functions
Logarithmic Functions
Approximate the function at the value of $x$ to four decimal places. (a) $f(x)=13^{\sqrt{x+1.1}}, \quad x=3$ (b) $g(x)=\left(\frac{5}{42}\right)^{-x}, \quad x=1.43$ (c) $h(x)=\left(2^{x}+2^{-x}\right)^{2 x}, \quad x=1.06$
Exponential Functions
Exer. 83-84: Approximate the function at the value of $x$ to four decimal places. (a) $f(x)=\ln (x+1)+e^{x}, \quad x=2$ (b) $g(x)=\frac{(\log x)^{2}-\log x}{4}, \quad x=3.97$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD