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3 years ago

Find the indicated functional value for the floor function: f(−0.75)f(−0.75)?

Mathematics

1 answer:

nikdorinn [45]3 years ago

6 0

We need to find the function value of floor function

Floor function comes under integer functions. we have floor and ceiling function.

Ceiling function gives the least integer value greater than or equal to x.

Floor function gives the greatest integer values less than or equal to x.

For example Celing(3.8) = 4

floor (3.8)= 3

Like that floor f(-0.75) = -1

greatest integer value less than -0.75 is -1

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3 years ago

Find the general expression for the slope of a line tangent to the curve of y=2x^2+4x at the point P(x,y) . Then find the slopes

ArbitrLikvidat [17]

Complete Question

Find the general expression for the slope of a line tangent to the curve of y=2x^2+4x at the point P(x,y) . Then find the slopes for $x = 3$ and x=0.5. Sketch the curve and the tangent lines. What is the general expression for the slope of a line tangent to the curve of the function y=2x^2+4 at the point P(x,y) ?

Answer:

The generally expression for the slope of $y = 2x^2 + 4x$ is $y' = 4x +4$

The graph is shown on the first uploaded image

The generally expression for the slope of $y=2x^2+4$ is $y' = 4x$

Step-by-step explanation:

From the question we are told that

The equation of the curve is $y = 2x^2 + 4x$

First we differentiate the equation

$y' = 4x +4$

Therefore the generally expression for the slope tangent to the curve $y=2x^2+4x$ is $y' = 4x +4$

The next step is to substitute for x = 3 and x = 0.5

So for $x_1 = 3$

$y' = 4(3) +4$

$y' =m_1= 16$

And for $x_2 = 0.5$

$y' = 4(0.5) +4$

$y' =m_2= 6$

Here m_1 and m_2 are slops of the curve

Next we obtain the coordinates of the tangent lines

So at $x_1 = 3$

$y_1 = 2(3)^2 + 4(3)$

$y_1 = 21$

So the coordinate for the first tangent line is

$(x_1 , y_1 ) = (3 , 21)$

At $x_2 = 0.5$

$y_2 = 2(0.5)^2 + 4(0.5)$

=> $y_2 = 2.5$

So the coordinate for the second tangent line is

$(x_2 , y_2 ) = (0.5 , 2.5)$

Next we obtain the equation for the tangent lines

So generally the slope is mathematically represented as

$m = \frac{y - y_1 }{x-x_1}$

For $(x_1 , y_1 ) = (-3 , 21)$ and $y' =m_1= 16$

$16 = \frac{y -21 }{x-3)}$

=> $y = 16x - 27$

For $(x_2 , y_2 ) = (0.5 , 2.5)$ and $y' =m_2= 6$

$6 = \frac{y -2.5 }{x-0.5}$

$y = 6x -0.5$

Generally the general expression for the slope of a line tangent to the curve of the function y=2x^2+4 at the point P(x,y) is mathematically evaluated by differentiating y=2x^2+4 as follows

$y' = 4x$

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