Logarithmic and Square Root Graphs

Question 1

Which of the following graphs represents $y = lnx?$

Your answer is correct.

This is $y = lnx .$

Not correct. Choice (b) is false.

Try again, this is $y = ex.$

Not correct. Choice (c) is false.

Try again, this is $√ -- y = x.$

Not correct. Choice (d) is false.

Try again, this is $y = ln(x+ 1).$ Remember $ln1 = 0 .$

Question 2

Which of the following graphs represents $√ -- y = x?$

Not correct. Choice (a) is false.

Try again, this is $y = ex.$

Not correct. Choice (b) is false.

Try again, this is $y = ln x.$

Your answer is correct.

This is $√-- y = x .$

Not correct. Choice (d) is false.

Try again, this is $√ ----- y = x+ 1.$
Remember √ -
0

= 0.

Question 3

What is the value of $f(x) = 5- 4 lnx$ at $x = 1?$

Your answer is correct.

Since $ln1 = 0$ we have $5 - 4 ln 1 = 5.$

Not correct. Choice (b) is false.

Try again, remember $ln1 = 0.$

Not correct. Choice (c) is false.

Try again, remember $ln1 = 0.$

Not correct. Choice (d) is false.

Try again, remember $ln 1 = 0.$

Question 4

What is the value of $f(x) = √x---5$ at $x = 1 ?$

Not correct. Choice (a) is false.

Try again, watch for negative numbers.

Not correct. Choice (b) is false.

Try again, $√ - 4-⁄= - 2 .$

Not correct. Choice (c) is false.

Try again, you have omitted the square root.

Your answer is correct.

The domain of this function is $[5,∞)$ since we cannot take the square root of a negative number.

Question 5

The following graphs represents $y = 3 + 2ln x, y = 3 + lnx ,$ $y = 3 + 2√x$ and $y = 3- 2lnx ,$ not necessarily in that order.
Which function matches which graph?

Not correct. Choice (a) is false.

Try again, you need to look at what happens at $x = 1$ and the steepness and directions of the graphs.

Your answer is correct.

Looking at the values of the functions at $x = 1$ we note that $f (1) ⁄= 3 4$ hence it must be $y = 3+ 2√x-.$
We note that $f 2$ has a negative gradient so it must be $3- 2lnx$
To determine which are $f 1$ and $f 2$ the steepest graph must be $3+ 2 lnx$ which makes $f = 3+ 2lnx 3$ and $f = 3+ lnx . 1$

Not correct. Choice (c) is false.

Try again, you need to look at what happens at $x = 1$ and the steepness and directions of the graphs.

Not correct. Choice (d) is false.

Try again, you need to look at the steepness of the graphs. You have $f1$ and $f3$ confused.


		Faculty of Economics and Business MathQuiz