22. Find the range of [tex]f(x)=\frac{x-2}{x-3}[/tex].
A. [tex]R \backslash{0,1}[/tex]
B. [tex]R[/tex]
C. [tex]R \backslash{2,3}[/tex]
D. [tex]R \backslash{1}[/tex].
23. The [tex]y[/tex]-intercept of [tex]y=a x^3+b x^2+c x+d[/tex] is
A. [tex](y(0), 0)[/tex]
B. [tex](y(0), y(0))[/tex]
C. [tex](0,0)[/tex]
D. [tex](0, y(0))[/tex].
24. Evaluate [tex]\lim _{h \rightarrow 0} \frac{1}{h} \ln \left(\frac{2+h}{2}\right)[/tex].
A. [tex]e^2[/tex]
B. 1
C. [tex]\frac{1}{2}[/tex]
D. 0.
25. Evaluate the left-hand limit [tex]\lim _{x \rightarrow-3}\left(-\frac{5}{x+3}\right)[/tex].
A. [tex]-x[/tex]
B. [tex]\propto[/tex]
C. 2
D. -2.
26. Evaluate [tex]\lim _{x \rightarrow-2} \frac{x+2}{\ln (x+3)}[/tex]
A. Does not exist
B. 2
C. 0
D. 1.
Answer (1)
Find the range of f ( x ) = x − 3 x − 2 by expressing x in terms of y and identifying the excluded value: R \ { 1 } .
Determine the y-intercept of y = a x 3 + b x 2 + c x + d by setting x = 0 : ( 0 , y ( 0 )) .
Evaluate lim h → 0 h 1 ln ( 2 2 + h ) using the derivative definition: 2 1 .
Find the left-hand limit lim x → − 3 ( − x + 3 5 ) , which approaches infinity: ∞ .
Evaluate lim x → − 2 l n ( x + 3 ) x + 2 using L'Hopital's rule: 1 .
Explanation
Introduction We are given five independent math problems to solve. We will address each one individually, providing a step-by-step solution for each.
Finding the Range Question 22: Find the range of f ( x ) = x − 3 x − 2 .
To find the range, we first express x in terms of y . Let y = x − 3 x − 2 . Then, y ( x − 3 ) = x − 2 , which gives y x − 3 y = x − 2 . Rearranging the terms, we have y x − x = 3 y − 2 , so x ( y − 1 ) = 3 y − 2 . Thus, x = y − 1 3 y − 2 .
The range consists of all possible values of y for which x is a real number. The only restriction is that the denominator y − 1 cannot be zero, so y = 1 . Therefore, the range is all real numbers except y = 1 , which is R \ { 1 } .
Finding the y-intercept Question 23: The y -intercept of y = a x 3 + b x 2 + c x + d is:
To find the y -intercept, we set x = 0 in the equation y = a x 3 + b x 2 + c x + d . This gives y = a ( 0 ) 3 + b ( 0 ) 2 + c ( 0 ) + d = d . The y -intercept is the point ( 0 , d ) , which can also be written as ( 0 , y ( 0 )) .
Evaluating the Limit Question 24: Evaluate lim h → 0 h 1 ln ( 2 2 + h ) .
We can rewrite the expression as lim h → 0 h l n ( 2 + h ) − l n ( 2 ) . This is the definition of the derivative of ln ( x ) evaluated at x = 2 . The derivative of ln ( x ) is x 1 , so the limit is 2 1 .
Evaluating the Left-Hand Limit Question 25: Evaluate the left-hand limit lim x → − 3 ( − x + 3 5 ) .
As x approaches − 3 from the left, x < − 3 , so x + 3 < 0 . Thus, x + 3 is a small negative number. Therefore, x + 3 1 is a large negative number. Then, − x + 3 5 is a large positive number. So, the limit is ∞ .
Evaluating the Limit Using L'Hopital's Rule Question 26: Evaluate lim x → − 2 l n ( x + 3 ) x + 2 .
Substituting x = − 2 into the expression, we get l n ( − 2 + 3 ) − 2 + 2 = l n ( 1 ) 0 = 0 0 , which is an indeterminate form. We can use L'Hopital's rule: x → − 2 lim ln ( x + 3 ) x + 2 = x → − 2 lim d x d ( ln ( x + 3 )) d x d ( x + 2 ) = x → − 2 lim x + 3 1 1 = x → − 2 lim ( x + 3 ) = − 2 + 3 = 1.
Summary of Answers Final Answers:
The range of f ( x ) = x − 3 x − 2 is R \ { 1 } .
The y -intercept of y = a x 3 + b x 2 + c x + d is ( 0 , y ( 0 )) .
lim h → 0 h 1 ln ( 2 2 + h ) = 2 1 .
lim x → − 3 ( − x + 3 5 ) = ∞ .
lim x → − 2 l n ( x + 3 ) x + 2 = 1 .
Examples
These problems cover fundamental concepts in calculus and algebra. Understanding ranges of functions is crucial in fields like economics, where you might want to know the possible output values of a cost function. Finding y-intercepts helps in modeling real-world phenomena, such as the initial value of a population growth model. Evaluating limits is essential in physics for calculating instantaneous rates of change, like velocity or acceleration. These skills collectively build a strong foundation for advanced mathematical modeling and problem-solving in various disciplines.
