Question
Find the maximum value of $f(x)=\left(\sqrt{-3+4 x-x^{2}}+4\right)^{2}+(x-5)^{2}$where $1 \leq x \leq 3$.
Step 1
The derivative of $f(x)$ is given by: \[f'(x) = 2(\sqrt{-3+4x-x^2}+4)\left(\frac{-2x+4}{2\sqrt{-3+4x-x^2}}\right) + 2(x-5)\] Show more…
Show all steps
Your feedback will help us improve your experience
Ankit Singh and 89 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
Find the maximum value of the function $$f(x)=3+4 x^{2}-x^{4}$$
Polynomial and Rational Functions
Quadratic Functions and Models
Find the $\max$ or min values of $f(x)=\sqrt{3 x^{2}-2 x^{3}}$
The Maxima and Minima
Level I
Find the maximum or minimum value of the function. $$f(x)=3-4 x-x^{2}$$
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD