Question
The function $f(x)=(\log (x-1))^{2}(x-1)^{2}$ has(a) local extremum at $x=1$(b) point of inflection at $x=1$(c) local extremum at $x=2$(d) point of inflection at $x=2$
Step 1
Using the chain rule and the product rule, we get \[f'(x) = 2(x-1)\log(x-1)(1+\log(x-1)) + 2(x-1)^2\frac{1}{x-1}.\] Show more…
Show all steps
Your feedback will help us improve your experience
Aman Gupta and 69 other Calculus 1 / AB 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
What value of $a$ makes $f(x)=x^{2}+(a / x)$ have \begin{equation}\begin{array}{l}{\text { a. } \text { a local minimum at } x=2 ?} \\ {\text { b. a point of inflection at } x=1 ?}\end{array}\end{equation}
Applications of Derivatives
Applied Optimization
What value of $a$ makes $f(x)=x^{2}+(a / x)$ have a. a local minimum at $x=2 ?$ b. a point of inflection at $x=1 ?$
Use the derivative of the function $y=f(x)$ to find the points at which $f$ has a (a) local maximum, (b) local minimum, or (c) point of inflection. $$y^{\prime}=(x-1)^{2}(x-2)$$
Connecting f and f with the Graph of f
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD