2 years ago

HELP! WHOEVER ANSWERS FIRST I'LL NAME BRAINLIAST!!!Part 1 Evaluate F(0)

1 answer:

3 0

The function that will be used when x = 0 is f(x) = x^2 as x = 0 fits the inequality condition of x ≤ 1.
The value of the function when x = 0 is f(0) = 0^2 = 0

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VikaD [51]

Answer:

11.25 // 11 $\frac{1}{4}$

Step-by-step explanation:

15×0.75

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dexar [7]

Answer:

$f^{-1}(x)=\frac{1}{2}+\sqrt{x-\frac{3}{4}}$

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y=x^2-x+1

We want to solve for x.

I'm going to use completing the square.

Subtract 1 on both sides:

y-1=x^2-x

Add (-1/2)^2 on both sides:

y-1+(-1/2)^2=x^2-x+(-1/2)^2

This allows me to write the right hand side as a square.

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Now remember we are solving for x so now we square root both sides:

$\pm \sqrt{y-3/4}=x-1/2$

The problem said the domain was 1/2 to infinity and the range was 3/4 to infinity.

This is only the right side of the parabola because of the domain restriction. We want x-1/2 to be positive.

That is we want:

$\sqrt{y-3/4}=x-1/2$

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$1/2+\sqrt{y-3/4}=x$

The last step is to switch x and y:

$1/2+\sqrt{x-3/4}=y$

$y=1/2+\sqrt{x-3/4}$

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The answer is amortization

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