Math 250 Self-grading assignment 7

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Best submitted over the internet by 3:00 PM on Wednesday 17 October 2018

Math 76O-250 Calculus for Business
Self-marking on-line assignment 7
Finding Critical Values and Candidates for Exponential and Logarithmic Functions
Best submitted over the internet by 3:00 PM on Wednesday 17 October 2018

Methods will be discussed in class

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Using a calculator is NOT allowed on this practice assignment.
The exercises below ask you to find critical values and candidates, but not to test them.

The term CRITICAL VALUE (of a function f(x) ) describes a value of x for which f '(x) = 0, or is undefined. Some texts use an alternate term "critical number or critical point".
The term CANDIDATE describes a value of x for which f "(x) = 0 or is undefined. Some texts use the phrase " second order critical value".
In an interactive tutorial elsewhere on this website, you can see several graphical examples of CRITICAL VALUES and CANDIDATES.
Important tools in locating CRITICAL VALUES and CANDIDATES are algebra rules and rules for differentiation.

Here is the solution of a more challenging problem, similar to problems [5] and [10] below:

Find all critical values of f(x) if f(x) = x

x³ (with x > 0). First differentiate using the quotient rule:

then f '(x) =

(1/x)x³ - 3x² ( x)

(x³)²
=

x² - 3x² ( x)

x⁶
=

1 - (3 x)

x⁴

Critical values will be values of x which make the above fraction equal to 0 .
The above fraction is 0 only if the numerator [1 - (3 x) ] = 0 .
Thus (3 x) = 1, and x = ⅓ .
Now raise the number "e" to both sides, to get [e ^x = e^⅓ ] or [ x = e^⅓ ].
Thus our desired critical value is e^⅓, which is about 1.3956
.
To find the one candidate for this function, differentiate the above f '(x) again: a bit messy but similar to above:
thus f "(x) =

(-3)(1/x)x⁴ - 4x³ (1 -3 x)

(x⁴)²
=

-7x³ + 12x³ ( x)

x⁸
=

-7 + 12( x)

x⁵

Candidates will be values of x which make the above fraction equal to 0 .
The above fraction is 0 only if the numerator [-7 + 12( x) ] = 0 .
Thus (12 x) = 7, and x = 7/12 (7 divided by 12) .
Now raise the number "e" to both sides, to get [e ^x = e^7/12 ] or [ x = e^7/12 ].
Thus our desired candidate is e^7/12, which is about 1.792 .

[1] (value 10%)	If f (x) = (x + 1) e^x ,
	then the one critical value of f is (enter a number)

[2] (value 10%)	If f (x) = (x - 1) e^{- x} ,
	then the one critical value of f is (enter a number)

[5] (value 10%)	If f (x) = x² x (for x > 0),
	then the one critical value of f is (enter a decimal accurate to 0.01: answer is between 0 and 1)

[6] (value 10%)	(see problem [1] above) If f (x) = (x + 1) e^x ,
	then the one candidate of f is (enter a number)

[7] (value 10%)	(see problem [2] above) If f (x) = (x - 1) e^{- x} ,
	then the one candidate of f is (enter a number)

[9] (value 10%)	If f (x) = e^x - e^{- x} ,
	then the one candidate of f is (enter a number)

[10] (value 10%)	(see problem 5) If f (x) = x² x (for x > 0),
	then the one candidate of f is (enter a decimal accurate to 0.01: answer is netween 0 and 1)

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