19. The inverse of [tex]f(x)=2 x-1[/tex] is
A. [tex]x-1[/tex]
B. [tex]\frac{x+1}{2}[/tex]
C. [tex]\frac{1}{x+2}[/tex]
D. [tex]x^2[/tex]
20. Find the inverse of [tex]f(x)=x^3-2[/tex].
A. [tex](x+2)^{\frac{1}{2}}[/tex]
B. [tex](x-2)^{\frac{1}{3}}[/tex]
C. [tex](x-2[/tex] [tex](x+2)[/tex].
21. If [tex]h(x)=\frac{x+1}{2 x-5}[/tex], find [tex]h^{-1}(0)[/tex].
A. o
B. -1
C. 1
D. [tex]\frac{\pi}{2}[/tex].
22. Find the range of [tex]f(x)=\frac{x-2}{x-3}[/tex], A, [tex]R \{0,1\}[/tex] B, R C, [tex]R \{2,3\}[/tex] D, [tex]R \[/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[/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 lett-hand limit [tex]\lim _{x \rightarrow-3}\left(-\frac{5}{x+3}\right)[/tex], A. [tex]-\infty[/tex] B. [tex]x[/tex] C. 2 D.
26. Evaluate [tex]\lim _{x \rightarrow-2} \frac{x+2}{\ln (x+3)}[/tex].
A. Does not exist
B. 2
C. o
D, 1 .
Answer (1)
Question 19: Find the inverse of f ( x ) = 2 x − 1 . The inverse is 2 x + 1 .
Question 20: Find the inverse of f ( x ) = x 3 − 2 . The inverse is ( x + 2 ) 3 1 .
Question 21: If h ( x ) = 2 x − 5 x + 1 , find h − 1 ( 0 ) . The value is − 1 .
Question 22: Find the range of f ( x ) = x − 3 x − 2 . The range is \boxed{R \{1}} .
Question 23: The y -intercept of y = a x 3 + b x 2 + c x + d is ( 0 , y ( 0 )) .
Question 24: Evaluate lim h → 0 h 1 ln ( 2 2 + h ) . The limit is 2 1 .
Question 25: Evaluate the left-hand limit lim x → − 3 ( − x + 3 5 ) . The limit is ∞ .
Question 26: Evaluate lim x → − 2 l n ( x + 3 ) x + 2 . The limit is 1 .
Explanation
Problem Overview We are given a set of multiple-choice questions and we need to select the correct answer for each one. Let's address them one by one.
Question 19 Solution Question 19: Find the inverse of f ( x ) = 2 x − 1 . To find the inverse, we set y = 2 x − 1 and solve for x in terms of y : y = 2 x − 1
y + 1 = 2 x
x = 2 y + 1 Now, we swap x and y to get the inverse function: f − 1 ( x ) = 2 x + 1 So the answer is B.
Question 20 Solution Question 20: Find the inverse of f ( x ) = x 3 − 2 . To find the inverse, we set y = x 3 − 2 and solve for x in terms of y : y = x 3 − 2
y + 2 = x 3
x = ( y + 2 ) 3 1 Now, we swap x and y to get the inverse function: f − 1 ( x ) = ( x + 2 ) 3 1 So the answer is C.
Question 21 Solution Question 21: If h ( x ) = 2 x − 5 x + 1 , find h − 1 ( 0 ) . To find h − 1 ( 0 ) , we first find the inverse function. Let y = 2 x − 5 x + 1 . Then, we solve for x in terms of y : y ( 2 x − 5 ) = x + 1
2 x y − 5 y = x + 1
2 x y − x = 5 y + 1
x ( 2 y − 1 ) = 5 y + 1
x = 2 y − 1 5 y + 1 So, h − 1 ( y ) = 2 y − 1 5 y + 1 . Now, we evaluate h − 1 ( 0 ) : h − 1 ( 0 ) = 2 ( 0 ) − 1 5 ( 0 ) + 1 = − 1 1 = − 1 So the answer is B.
Question 22 Solution Question 22: Find the range of f ( x ) = x − 3 x − 2 . The function has a vertical asymptote at x = 3 . To find the horizontal asymptote, we examine the limit as x approaches infinity: x → ∞ lim x − 3 x − 2 = 1 Thus, the horizontal asymptote is y = 1 . The range is all real numbers except y = 1 . So the answer is A. R \{1}
Question 23 Solution Question 23: The y -intercept of y = a x 3 + b x 2 + c x + d is found by setting x = 0 : y ( 0 ) = a ( 0 ) 3 + b ( 0 ) 2 + c ( 0 ) + d = d The y -intercept is the point ( 0 , d ) , which is ( 0 , y ( 0 )) . So the answer is D.
Question 24 Solution Question 24: Evaluate lim h → 0 h 1 ln ( 2 2 + h ) . We can rewrite this as: h → 0 lim h ln ( 2 + h ) − ln ( 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 . So the answer is C.
Question 25 Solution 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 . Therefore, x + 3 approaches 0 from the negative side. Thus, − x + 3 5 approaches − 0 − 5 , which is + ∞ . So the answer is B. ∞
Question 26 Solution 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 So the answer is D.
Examples
Understanding limits and derivatives is crucial in physics for calculating instantaneous velocity and acceleration. For example, if you have a function describing the position of a car over time, finding the derivative at a specific time gives you the car's instantaneous velocity at that moment. Similarly, finding the limit of a function helps in understanding the behavior of physical systems as they approach certain conditions, such as the behavior of an electrical circuit as the resistance approaches zero.
