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4 years ago

How do you know when a function is increasing by looking at its graph

2 answers:

7 0

If f′(x) > 0, then f is increasing on the interval, and if f′(x) < 0, then f is decreasing on the interval. This and other information may be used to show a reasonably accurate sketch of the graph of the function. Example 1: For f(x) = x 4 − 8 x 2 determine all intervals where f is increasing or decreasing.

fredd [130]4 years ago

3 0

The way to tell if a function is increasing via its graph is by seeing if the value of y is increasing as the value of x increases, which tends to look like the line heading towards the top right of the plane.

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Find the area of the given figure. Show your work!

Svetlanka [38]

The area of the figure above is 53

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3 years ago

Delbert Rowell drove his tractor rig 58,125 miles in 6 months. Estimate how many miles he drove in each month

zzz [600]

Answer:

just do 58,125 divided by 6

Step-by-step explanation:

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4 years ago

Someone help me fast its due at 8 pm

djverab [1.8K]

Answer:

Eric's statement is false.

Step-by-step explanation:

When a positive and a negative number is added, pay attention to what number has a greater absolute value. If the positive number is greater, then the answer will be positive. In the negative number is greater, then the answer will be negative. For example, 23 + (-4) is going to end up as a positive sum, since 23 has a greater absolute value than -4. On the other hand, (-23) + 4 is going to end up as a negative number since -23 has a greater absolute value than 4.

5 0

3 years ago

Consider the function. f(x)= -2/3x-24 Which conclusions can be drawn about f–1(x)? Select two options. f–1(x) has a slope of -2/

disa [49]

Answer:

(D) $f^{-1}{(x$) has an x-intercept of (-36, 0).$

(E) $f^{-1}{(x)$ has a range of all real numbers.

Step-by-step explanation:

Given the function: $f(x)= -\dfrac23x-24$

$f(x)+24= -\dfrac23x\\$Multiply both sides by $ -\dfrac32\\x=-\dfrac32(f(x)+24)\\f^{-1}(x)=-\dfrac32f(x)-36$

When y=f(x)=0

$f^{-1}(x)=-\dfrac32(0)-36\\f^{-1}(x)=-36$

Therefore, $f^{-1}(x)$ has a x-intercept of -36$

Also, $f^{-1}(x)$ has a range of all real numbers.

Therefore, Options D and E are correct.

5 0

3 years ago

1. Approximate the given quantity using a Taylor polynomial with n3.

Jet001 [13]

Answer:

See the explanation for the answer.

Step-by-step explanation:

Given function:

$f(x) = x^{1/4}$

The n-th order Taylor polynomial for function f with its center at a is:

$p_{n}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(n)}a}{n!} (x-a)^{n}$

As n = 3 So,

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{3!} (x-a)^{3}$

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{6} (x-a)^{3}$

$p_{3}(x) = a^{1/4} + \frac{1}{4a^{ 3/4} } (x-a)+ (\frac{1}{2})(-\frac{3}{16a^{7/4} } ) (x-a)^{2} + (\frac{1}{6})(\frac{21}{64a^{11/4} } ) (x-a)^{3}$

$p_{3}(x) = 81^{1/4} + \frac{1}{4(81)^{ 3/4} } (x-81)+ (\frac{1}{2})(-\frac{3}{16(81)^{7/4} } ) (x-81)^{2} + (\frac{1}{6})(\frac{21}{64(81)^{11/4} } ) (x-81)^{3}$

$p_{3} (x)$ = 3 + 0.0092592593 (x - 81) + 1/2 ( - 0.000085733882) (x - 81)² + 1/6

(0.0000018522752) (x-81)³

$p_{3} (x)$ = 0.0092592593 x - 0.000042866941 (x - 81)² + 0.00000030871254

(x-81)³ + 2.25

Hence approximation at given quantity i.e.

x = 94

Putting x = 94

$p_{3} (94)$ = 0.0092592593 (94) - 0.000042866941 (94 - 81)² +

0.00000030871254 (94-81)³ + 2.25

= 0.87037 03742 - 0.000042866941 (13)² + 0.00000030871254(13)³ +

2.25

= 0.87037 03742 - 0.000042866941 (169) +

0.00000030871254(2197) + 2.25

= 0.87037 03742 - 0.007244513029 + 0.0006782414503 + 2.25

$p_{3} (94)$ = 3.113804102621

Compute the absolute error in the approximation assuming the exact value is given by a calculator.

Compute $\sqrt[4]{94}$ as $94^{1/4}$ using calculator

Exact value:

$E_{a}$ (94) = 3.113737258478

Compute absolute error:

Err = | 3.113804102621 - 3.113737258478 |

Err (94) = 0.000066844143

If you round off the values then you get error as:

|3.11380 - 3.113737| = 0.000063

Err (94) = 0.000063

If you round off the values up to 4 decimal places then you get error as:

|3.1138 - 3.1137| = 0.0001

Err (94) = 0.0001

4 0

4 years ago