Estimate to the nearest 0.5 unit and classify the extrema for the graph of each function.

Mathematics

1 answer:

VLD [36.1K]3 years ago

3 0

$f(x) = x^3+x^2-x$

Finding the extreme point :

$f'(x)=3x^2+2x-1=0\\\\3x^2+3x-x-1=0\\\\3x(x+1)-1(x+1)=0\\\\(3x-1)(x+1)=0\\\\x=\dfrac{1}{3}\ ,\ x=-1$

Now,

f''( x ) at x = 1/3 and -1 is :

$f''(x)=6x+2\\\\f''(1/3)=4 \\\\f''(-1)=-4$

Therefore, it is maximum at x = -1 and minimum at x = 1/3 .

Hence, this is the required solution.

You might be interested in

A brewery has a beer dispensing machine that dispenses beer into the company's 12 ounce bottles. The distribution for the amount

-Dominant- [34]

Answer:

The company should use a mean of 12.37 ounces.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The distribution for the amount of beer dispensed by the machine follows a normal distribution with a standard deviation of 0.17 ounce.

This means that $\sigma = 0.17$

The company can control the mean amount of beer dispensed by the machine. What value of the mean should the company use if it wants to guarantee that 98.5% of the bottles contain at least 12 ounces (the amount on the label)?

This is $\mu$ , considering that when $X = 12$ , Z has a p-value of $1 - 0.985 = 0.015$ , so when $X = 12, Z = -2.17$ .