Consider the function.
f(x)=2 1/2 - 3 1/3
What is the x-intercept of f-1(x)?

o (-3 1/3,0)
o (-3/10,0)
o (3/4,0)
o (2 1/2,0)

Consider The Function.f(x)=2 1/2 - 3 1/3What Is The X-intercept Of F-1(x)?o (-3 1/3,0)o (-3/10,0)o (3/4,0)o

Answers

Answer 1

Answer:

Option 4

Step-by-step explanation:

Given function is,

[tex]f(x)=2\frac{1}{2}-3\frac{1}{3}x[/tex]

Steps to get the inverse of the given function,

Step 1,

Convert the function into equation,

[tex]y=2\frac{1}{2}-3\frac{1}{3}x[/tex]

Step 2,

Interchange the variables 'x' and 'y',

[tex]x=2\frac{1}{2}-3\frac{1}{3}y[/tex]

Step 3,

Solve the equation for the value of y,

[tex]x-2\frac{1}{2}=-3\frac{1}{3}y[/tex]

[tex]x-\frac{5}{2}=-\frac{10}{3}y[/tex]

[tex]-x+\frac{5}{2}=\frac{10}{3}y[/tex]

[tex]-3x+\frac{15}{2}=10y[/tex]

[tex]-\frac{3}{10}x+\frac{15}{20}=y[/tex]

[tex]y=-\frac{3}{10}x+\frac{3}{4}[/tex]

Step 4,

Convert the equation into inverse function,

[tex]f^{-1}x=-\frac{3}{10}x+\frac{3}{4}[/tex]

For x-intercepts,

Substitute y = 0,

[tex]0=-\frac{3}{10}x+\frac{3}{4}[/tex]

[tex]\frac{3}{10}x=\frac{3}{4}[/tex]

[tex]x=\frac{10}{4}[/tex]

[tex]x=\frac{5}{2}[/tex]

[tex]x=2\frac{1}{2}[/tex]

Therefore, x-intercept of the inverse function is [tex](2\frac{1}{2},0)[/tex].

Option 4 will be the answer.


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Consider the function f(x)=x−5ln(x), 1/5≤x≤10.
The absolute maximum value is =

and this occurs at x equals =

The absolute minimum value is =

and this occurs at x equals =

Answers

f(x) = x - 5 ln(x)

Get the derivative:

f'(x) = 1 - 5/x

Find the critical points:

• f'(x) = 1 - 5/x = 0   ->   5/x = 1   ->   x = 5

• f'(x) is undefined for x = 0, but that's outside the domain of f(x)

Get the second derivative:

f''(x) = 5/x ^2

Check the sign of the second derivative at the critical point x = 5:

f'' (5) = 5/5^2 = 5/25 = 1/5 > 0

Since the second derivative is positive at this point, this means x = 5 is a local minimum.

Check the value of f(x) at the critical point as well as the endpoints of the given domain:

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So we have

• absolute maximum = 1/5 - 5 ln(1/5) at x = 1/5

• absolute minimum = 5 - 5 ln(5) at x = 5

The absolute maximum value is  f(5)=5−5ln(5), and this occurs at x equals 5

The absolute minimum value is  f(1/5)= 1/5−5ln(1/5), and this occurs at x equals 1/5

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Substitute the value of x = 5 into the second derivative function to have:

f''(5) = -5/5² = 1/5

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The absolute maximum value is  f(5)=5−5ln(5), and this occurs at x equals 5

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Answer:

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When I try to type the right answer it says incorrect answer

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