PLEASE HURRY!!!!! What is the turning point of the graph of f(x) = |x – 4| ?

1 answer:

8 0

$f(x)=|x-4|\\
f'(x)=\text{sgn} (x-4)\\
\text{sgn} (x-4)=0\\
x-4=0\\
x=4\\\\

$

For $x$ $f'(x)$ .
For $x>4$ $f'(x)>0$ .
$\Downarrow\\
$
At $x=4$ there's turning point, which is equal $|4-4|=|0|=0$ .

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H(x) = x3 + 7<br> Determine whether the function is one-to-one.

zubka84 [21]

Answer:

we conclude that the function is one-to-one.

Step-by-step explanation:

A function will a one-to-one function if it

passes the vertical line test to make sure it is indeed a function, and

also a horizontal line test to make sure it is it one-to-one.

In other words,

The function will be one-to-one if it passes the vertical line test, and also if the horizontal line only cuts the graph of the function in one place.

The reason is that there must be only one x-value for each y-value.

Given the function

$h\left(x\right)\:=\:x^3\:+\:7$

Have a look at the attached graph.

The red portion represents the graph of the function $h\left(x\right)\:=\:x^3\:+\:7$ .

The green portion represents the graph of x=2 which is basically a vertical line test. Vertical line indicates that it cuts the cuts the graph of the function in one place. So it is clear that $h\left(x\right)\:=\:x^3\:+\:7$ is indeed a function.

The blue line represents the graph of y=9, which is basically a horizontal line test. Horizontal line indicates that it cuts the cuts the graph of the function in one place. So it is clear that $h\left(x\right)\:=\:x^3\:+\:7$ is a one-to-one function, as there is only one x-value for each y-value.

Therefore, we conclude that the function is one-to-one.