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klio [65]
3 years ago
5

Let $f(x) = x^2 + 4x - 31$. for what value of $a$ is there exactly one real value of $x$ such that $f(x) = a$?

Mathematics
1 answer:
enot [183]3 years ago
7 0

To solve for this, we need to find for the value of x when the 1st derivative of the equation is equal to zero (or at the extrema point).

So what we have to do first is to derive the given equation:

f (x) = x^2 + 4 x – 31

 

Taking the first derivative f’ (x):

f’ (x) = 2 x + 4

 

Setting f’ (x) = 0 and find for x:

2 x + 4 = 0

x = - 2

 

Therefore the value of a is:

a = f (-2)

a = (-2)^2 + 4 (-2) – 31

a = 4 – 8 – 31

a = - 35

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Step-by-step explanation:

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4 0
3 years ago
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What is the inverse function of f(x) = 25x-3?
Harrizon [31]

Answer:

The answer is

<h2>{f}^{ - 1} (x) =  \frac{x +3 }{25}</h2>

Step-by-step explanation:

f(x) = 25x - 3

To find the inverse of f(x) , equate f(x) to y

That's

y = f(x)

We have

y = 25x - 3

Next interchange the terms that's x becomes y and y becomes x

x = 25y - 3

<u>Next solve for y</u>

Move 3 to the other side of the equation

That's

25y = x +  3

<u>Divide both sides by 25</u>

\frac{25y}{25}  =  \frac{x + 3}{25}

We have the final answer as

{f}^{ - 1} (x) =  \frac{x +3 }{25}

Hope this helps you

5 0
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natta225 [31]

Answer:

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Step-by-step explanation:

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