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Calculus 1 : Local Minimum

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Example Questions

Example Question #1502 : Functions

Find the local minimum of $\displaystyle f(x)=4x^2-x$.

Possible Answers:

$x=-\frac{1}{8}$ $\displaystyle x=-\frac{1}{8}$

$x=\frac{1}{8}$ $\displaystyle x=\frac{1}{8}$

x=0 $\displaystyle x=0$

x=2 $\displaystyle x=2$

Correct answer:

$x=\frac{1}{8}$ $\displaystyle x=\frac{1}{8}$

Explanation:

First, take the derivative of the function, remembering to multiply the exponent by the coefficeint and then subtracting one from the exponent. Therefore, the derivative is: $\displaystyle f{}'(x)=8x-1$. Then, set it equal to 0 to get your critical point: $8x-1=0; x=\frac{1}{8}$ $\displaystyle 8x-1=0; x=\frac{1}{8}$. Then, test a value on either side of your critical point and plug into the derivative. To the left, the value is negative. To the right, the value is positive. Therefore, your local minimum is at $x=\frac{1}{8}$ $\displaystyle x=\frac{1}{8}$.

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Example Question #52 : Curves

Given the function $\displaystyle y=5x^2+2x$, find its local minimum.

Possible Answers:

$x=-\frac{1}{5}$ $\displaystyle x=-\frac{1}{5}$

x=-5 $\displaystyle x=-5$

x=0 $\displaystyle x=0$

x=5 $\displaystyle x=5$

Correct answer:

$x=-\frac{1}{5}$ $\displaystyle x=-\frac{1}{5}$

Explanation:

To find the local minimum of a function, one must find the critical points (where the derivative is equal to zero).

Taking the derivative of the function above using the power rule,

$\frac{\mathrm{d} }{\mathrm{d} x}x^n= nx^{n-1}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n= nx^{n-1}$.

The derivative is

$y'=2\cdot 5x^{2-1}+1\cdot 2x^{1-1}$ $\displaystyle y'=2\cdot 5x^{2-1}+1\cdot 2x^{1-1}$

$\displaystyle y'(x)= 10x+2$.

Setting this equal to zero and solving for x gets you

$\\0=10x+2 \\ \\ \frac{-2}{10}=\frac{10x}{10} \\ \\-\frac{1}{5}=x$ $\displaystyle \\0=10x+2 \\ \\ \frac{-2}{10}=\frac{10x}{10} \\ \\-\frac{1}{5}=x$

To see if this indeed is a local minimum one must plug in values below and above this value of x to see if the derivative is positive (increasing) or negative (decreasing) around that point.

If you plug -2 into the equation for y'(x) $\displaystyle y'(x)$ you get -18 (decreasing), and plugging in 0, you get $\displaystyle y'(x)= 2$ (increasing).

Thus, $x=-\frac{1}{5}$ $\displaystyle x=-\frac{1}{5}$ is a local minimum.

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Calculus 1 : Local Minimum

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1502 : Functions

Example Question #52 : Curves

All Calculus 1 Resources

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