Given:
The graph of a function is given.
To find:
The range of the graph.
Solution:
We know that, the domain is the set of input values and range is the set of output values.
In a graph, domain is represented by the x-axis and range is represented by the y-axis.
From the given graph it is clear that there is an open circle at (-8,-8) and a closed circle at (3,4). It means the function is not defined at (-8,-8) but defined for (3,4).
The graph of the function is defined over the interval
. So, the domain is (-8,3].
The values of the function lie in the interval
. So, the range is (-8,4].
Therefore, the range of the function are all real values over the interval (-8,4].
2
42 ÷ 2 = 21
16 ÷ 2 = 8
Let's try the other factors of 42...
42 ÷ 6 = 7
16 ÷ 6 = 2.66
No...
42 ÷ 7 = 6
16 ÷ 7 = 2 2/7
No...
42 ÷ 14 = 3
16 ÷ 14 = 1.1428
No...
Answer:
<u>y'= 5x^4 + 5^x In(5)</u>
Step-by-step explanation:
<u>Differentiate</u><u> </u><u>with </u><u>Respect</u><u> </u><u>to</u><u> </u><u>x</u>
<u>f(</u><u>x)</u><u>'</u><u>=</u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>In(</u><u>5</u><u>^</u><u>x</u><u>)</u>
<u>f(</u><u>x)</u><u>'</u><u>=</u><u> </u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>x </u><u>In(</u><u>5</u><u>)</u>
<u>with </u><u>respect</u><u> </u><u>to </u><u>x,</u><u> </u><u>we </u><u>have</u>
<u>y'=</u><u> </u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>y </u><u>In(</u><u>5</u><u>)</u>
<u>y'=</u><u> </u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>5</u><u>^</u><u>x</u><u> </u><u>In(</u><u>5</u><u>)</u>
Im not so sure but, 3x2=6; 2x3=6; 6+6=12