Precalculus : Determine Points of Inflection

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #132 : Derivatives

Find the points of inflection for the following function:

\(\displaystyle f(x)=x^2-16x+100\)

Possible Answers:

\(\displaystyle 4\)

\(\displaystyle 16\)

\(\displaystyle 100\)

\(\displaystyle none\)

Correct answer:

\(\displaystyle none\)

Explanation:

To solve, find when the 2nd derivative is equal to 0 and plug it into the first equation. We must use the power rule, as outlined below, to find each derivative.

Power rule: \(\displaystyle (x^n)'=nx^{n-1}\)

Thus,

\(\displaystyle f(x)=x^2-16x+100\)

\(\displaystyle f'(x)=2x^{2-1}-16x^{1-1}=2x-16x^0=2x-16\)

\(\displaystyle f''(x)=1*2x^{1-1}=2x^0=2\)

Since 2 is never equal to 0, there are no points of inflection.

Example Question #21 : Determine Points Of Inflection

Determine the \(\displaystyle x\) values for the points of inflection of the following function:

\(\displaystyle f(x)=x^3-x^2+x-1\)

Possible Answers:

\(\displaystyle -2\)

\(\displaystyle 3\)

\(\displaystyle \frac{-1}{3}\)

\(\displaystyle \frac{1}{3}\)

Correct answer:

\(\displaystyle \frac{1}{3}\)

Explanation:

To solve, you must set the second derivative equal to 0 and solve for x. To differentiate twice, use the power rule as outlined below:

Power Rule: \(\displaystyle (x^n)'=nx^{n-1}\)

Therefore:

\(\displaystyle f'(x)=3x^{3-1}-2x^{2-1}+x^{1-1}=3x^2-2x\)

Remember, the derivative a constant is 0.

\(\displaystyle f''(x)=3*2x^{2-1}-2x^{1-1}=6x-2\)

Now, set it equal to 0. Thus,

\(\displaystyle 6x-2=0\)

\(\displaystyle 6x=2\)

\(\displaystyle x=\frac{2}{6}\)

\(\displaystyle x=\frac{1}{3}\)

 

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