# Find the critical points of the function f(x) = x^2 + \ln x on the interval [5,3]

## Question:

Find the critical points of the function {eq}f(x) = x^2 + \ln x {/eq} on the interval {eq}[5,3] {/eq}

## Critical points:

The critical points of a function are those points where the derivative of the function is zero i.e. the slope of the tangent at these points is zero and parallel to the x-axis.

## Answer and Explanation:

We have :

{eq}f(x) = x^2 + \ln x {/eq}

Now finding its derivative :

{eq}\displaystyle f'(x) = \frac{d}{dx}(x^2 + \ln x) \\\displaystyle f'(x) = 2x + \frac{1}{x} {/eq}

For critical points this derivative is zero,

{eq}\\\displaystyle f'(x) = 0 \\\displaystyle 2x + \frac{1}{x} = 0 \\\displaystyle \frac{2x^2+1}{x} = 0 {/eq}

Therefore,

{eq}2x^2+1 = 0 \\x = \sqrt{\frac{-1}{2}} {/eq}

Therefore critical points don't exist for this function in the interval {eq}[5,3] {/eq}

#### Learn more about this topic: Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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