# What are the points of inflection of #f(x)=8x - 4 lnx #?

The function has no points of inflection.

graph{8x-4lnx [-9.45, 56.4, 91.34, -40.3]}

By signing up, you agree to our Terms of Service and Privacy Policy

To find the points of inflection of ( f(x) = 8x - 4 \ln(x) ), you need to determine where the second derivative changes sign.

First, find the second derivative: [ f'(x) = 8 - \frac{4}{x} ] [ f''(x) = \frac{4}{x^2} ]

Next, set ( f''(x) = 0 ) to find any possible points of inflection: [ \frac{4}{x^2} = 0 ] [ x^2 \neq 0 ]

Since the second derivative is never equal to zero, there are no points of inflection for this function.

By signing up, you agree to our Terms of Service and Privacy Policy

- If #( x+2) / x#, what are the points of inflection, concavity and critical points?
- How do you find all critical point and determine the min, max and inflection given #f(x)=x^3+x^2-x#?
- For what values of x is #f(x)= x^2-x + 1/x-1/x^2 # concave or convex?
- How do you find all points of inflection given #y=x^3-2x^2+1#?
- How do you find the second derivative of #sqrtx#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7