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3 years ago

F(x)=4x2-3x+5 f(2) find function value

Mathematics

1 answer:

Dovator [93]3 years ago

8 0

Hello!

This question is an example of function notation. Function notation is basically f(x) , said as "f of x". To solve this problem, you need to substitute x = 2 into each x-value in this equation. F(2) is written as function notation, while x = 2 is not.

F(2) = 4x² - 3x + 5

F(2) = 4(2)² - 3(2) + 5

F(2) = 4(4) - 6 + 5

F(2) = 16 - 6 + 5

F(2) = 15

Therefore, the value of F(2) is 15.

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3 years ago

What would be the midpoint of a line segment with endpoints at (0,1)<br> and <br>(5,7) ?

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Answer:

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Step-by-step explanation:

Using the midpoint formula

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3 years ago

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3 years ago

Solve for x in the equation x squared + 11 x + StartFraction 121 Over 4 EndFraction = StartFraction 125 Over 4 EndFraction.

Ira Lisetskai [31]

Answer:

Below

Step-by-step explanation:

● x^2 + 11x + 121/4 = 125/4

Substract 125/4 from both sides:

● x^2 + 11x + 121/4-125/4= 125/4 -125/4

● x^2 + 11x - (-4/4) = 0

● x^2 +11x -(-1) = 0

● x^2 + 11 x + 1 = 0

This is a quadratic equation so we will use the determinanant (b^2-4ac)

● a = 1

● b = 11

● c = 1

● b^2-4ac = 11^2-4*1*1 = 117

So this equation has two solutions:

● x = (-b -/+ √(b^2-4ac) ) / 2a

● x = (-11 -/+ √(117) ) / 2

● x = (-11 -/+ 3√(13))/ 2

● x = -0.91 or x = -10.9

Round to the nearest unit

● x = -1 or x = -11

The solutions are { -1,-11}

6 0

3 years ago

Read 2 more answers

please show on graph (with x and y coordinates) state where the function x^4-36x^2 is non-negative, increasing, concave up

babunello [35]

Answer:

$y'' =12x^2 -72=0$

And solving we got:

$x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}$

We can find the sings of the second derivate on the following intervals:

$(-\infty$ Concave up

$x=-\sqrt{6}, y =-180$ inflection point

$(-\sqrt{6}$ Concave down

$x=\sqrt{6}, y=-180$ inflection point

$(\sqrt{6}$ Concave up

Step-by-step explanation:

For this case we have the following function:

$y= x^4 -36x^2$

We can find the first derivate and we got:

$y' = 4x^3 -72x$

In order to find the concavity we can find the second derivate and we got:

$y'' = 12x^2 -72$

We can set up this derivate equal to 0 and we got:

$y'' =12x^2 -72=0$

And solving we got:

$x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}$

We can find the sings of the second derivate on the following intervals:

$(-\infty$ Concave up

$x=-\sqrt{6}, y =-180$ inflection point

$(-\sqrt{6}$ Concave down

$x=\sqrt{6}, y=-180$ inflection point

$(\sqrt{6}$ Concave up

8 0

3 years ago