Concept explainers
(a).
To find: The real values of x that cause the function to be zero.
The real value of x are equal to −52,3 .
Given information:
Consider the given function.
f(x)=(2x+5)√x−3(x−4)2
Calculation:
When the given function f(x) is zero, then the values of x can be calculated as follow:
(2x+5)√x−3=02x+5=02x=−5x=−52
And
(2x+5)√x−3=0√x−3=0x−3=0x=3
Thus, the real values of x are equal to −52,3 .
(b).
To find: The real values of x that cause the function to be undefined.
The real values of x is equal to 4 .
Given information:
Consider the given function.
f(x)=(2x+5)√x−3(x−4)2
Calculation:
When the given function f(x) is undefined, then substitute the denominator equal to zero and calculate the values of x below:
(x−4)2=0x−4=0x=4
Therefore, the real values of x is 4 for the function f(x) is undefined.
(c).
To find: The real values of x that cause the function to be positive.
The real values of x is equal to (3,4)∪(4,∞) .
Given information:
Consider the given function.
f(x)=(2x+5)√x−3(x−4)2
Calculation:
The given function f(x) is positive.
First find the boundary points of the function f(x) .
Substitute the numerator and the denominator equal to zero, then calculate the values of x below:
(2x+5)√x−3=0x=−52,3
And
(x−4)2=0x=4
So, the boundary points of the function f(x) are −52,3,4 .
Now, locate these boundary points on the number line (or sign chart) and dividing the number line into intervals and check the function f(x) is positive or negative at each interval below:
Therefore, the given function f(x) is positive into (3,4)∪(4,∞) .
Hence, the real values of x is equal to (3,4)∪(4,∞) .
(d).
To find: The real values of x that cause the function to be negative.
No real values of x .
Given information:
Consider the given function.
f(x)=(2x+5)√x−3(x−4)2
Calculation:
The given function f(x) is negative.
First find the boundary points of the function f(x) .
Substitute the numerator and the denominator equal to zero, then calculate the values of x below:
(2x+5)√x−3=0x=−52,3
And
(x−4)2=0x=4
So, the boundary points of the function f(x) are −52,3,4 .
Now, locate these boundary points on the number line (or sign chart) and dividing the number line into intervals and check the function f(x) is positive or negative at each interval below:
Therefore, no value of x where the given function f(x) is negative.
Hence, there is no real values of x .
Chapter 2 Solutions
PRECALCULUS:GRAPHICAL,...-NASTA ED.
- H.w WI M Wz A Sindax Sind dy max Утах at 0.75m from A w=6KN/M L=2 W2=9 KN/m P= 10 KN B Make the solution handwritten and not artificial intelligence because I will give a bad rating if you solve it with artificial intelligencearrow_forwardSolve by DrWz WI P L B dy Sind Ⓡ de max ⑦Ymax dx Solve by Dr ③Yat 0.75m from A w=6KN/M L=2 W2=9 kN/m P= 10 KN Solve By Drarrow_forwardHow to find the radius of convergence for the series in the image below? I'm stuck on how to isolate the x in the interval of convergence.arrow_forward
- Determine the exact signed area between the curve g(x): x-axis on the interval [0,1]. = tan2/5 secx dx andarrow_forwardSet up the partial fraction expansion of the function below. Do not explicitly solve for the variables 5 x²(x − 2)(x − 3)³ (24 - 81)² -arrow_forwardEvaluate the integral below: (4w (4w8) sec(4w) tan(4w) dwarrow_forward
- solve these pleasearrow_forwardA factorization A = PDP 1 is not unique. For A= 7 2 -4 1 1 1 5 0 2 1 one factorization is P = D= and P-1 30 = Use this information with D₁ = to find a matrix P₁ such that - -1 -2 0 3 1 - - 1 05 A-P,D,P P1 (Type an integer or simplified fraction for each matrix element.)arrow_forwardMatrix A is factored in the form PDP 1. Use the Diagonalization Theorem to find the eigenvalues of A and a basis for each eigenspace. 30 -1 - 1 0 -1 400 0 0 1 A= 3 4 3 0 1 3 040 3 1 3 0 0 4 1 0 0 003 -1 0 -1 Select the correct choice below and fill in the answer boxes to complete your choice. (Use a comma to separate vectors as needed.) A basis for the corresponding eigenspace is { A. There is one distinct eigenvalue, λ = B. In ascending order, the two distinct eigenvalues are λ₁ ... = and 2 = Bases for the corresponding eigenspaces are { and ( ), respectively. C. In ascending order, the three distinct eigenvalues are λ₁ = = 12/2 = and 3 = Bases for the corresponding eigenspaces are {}, }, and { respectively.arrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning





