Ask question

Ask question

All categories

anastassius [24]

3 years ago

8

F(x)=4x^2+1 and g(x)=x^2-5 fins (f+g)(x)

2 answers:

larisa [96]3 years ago

6 0

Hi there!

$(f +g )(x) = f(x) +g (x)$
In other words: to find (f + g)(x) we must add up the functions f(x) and g(x). Substitute both functions into the formula.

$(f + g)(x) =( 4x {}^{2} + 1) + ( {x}^{2} - 5)$
Remove the parenthesis, since they don't have a function in this situation.

$(f + g)(x) =4x {}^{2} + 1 + {x}^{2} - 5$
Rearrange the expression in order to collect terms later.

$(f + g)(x) = 4x {}^{2} + {x}^{2} + 1 - 5$
And finally collect the terms.

$(f + g)(x) = 5x {}^{2} - 4$
~ Hope this helps you!

xenn [34]3 years ago

5 0

$(f+g)(x)=4x^2+1+x^2-5=5x^2-4$

You might be interested in

Consider the linear function that is represented by the equation y = 2 x + 2 and the linear function that is represented by the

Arisa [49]

Answer:

A

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x > 0)

vfiekz [6]

Answer:

Step-by-step explanation:

given a point $(x_0,y_0)$ the equation of a line with slope m that passes through the given point is

$y-y_0 = m(x-x_0)$ or equivalently

$y = mx+(y_0-mx_0)$ .

Recall that a line of the form $y=mx+b$ , the y intercept is b and the x intercept is $\frac{-b}{m}$ .

So, in our case, the y intercept is $(y_0-mx_0)$ and the x intercept is $\frac{mx_0-y_0}{m}$ .

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph $(x_0,\frac{1}{x_0})$ . Which means that $y_0=\frac{1}{x_0}$

The slope of the tangent line is given by the derivative of the function evaluated at $x_0$ . Using the properties of derivatives, we get

$y' = \frac{-1}{x^2}$ . So evaluated at $x_0$ we get $m = \frac{-1}{x_0^2}$

Replacing the values in our previous findings we get that the y intercept is

$(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}$

The x intercept is

$\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0$

The triangle in consideration has height $\frac{2}{x_0}$ and base $2x_0$ . So the area is

$\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2$

So regardless of the point we take on the graph, the area of the triangle is always 2.

6 0

3 years ago

PLS HELP ME ASAP FOR 3!!! (MUST SHOW WORK!!) + LOTS OF POINTS!! *brainliest*

Roman55 [17]

$2, because if you do 8*25 you get 200 and if you divide that by 100 you get a total of 2 making the tax on the purchase $2.

4 0

3 years ago

Read 2 more answers

Please help. It’s due today

vladimir1956 [14]

Answer:

There's nothing there I believe you forgot to add a link just add or create another question and i'll see what I can do :)

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Convert the Cartesian equation (x 2 + y 2)2 = 4(x 2 - y 2) to a polar equation.

SashulF [63]

<u>ANSWER</u>

${r}^{2} = 4 \cos2\theta$

<u>EXPLANATION</u>

The Cartesian equation is

${( {x}^{2} + {y}^{2} )}^{2} = 4( {x}^{2} - {y}^{2} )$

We substitute

$x = r \cos( \theta)$

$y = r \sin( \theta)$

and

${x}^{2} + {y}^{2} = {r}^{2}$

This implies that

${( {r}^{2} )}^{2} = 4(( { r \cos\theta) }^{2} - {(r \sin\theta) }^{2} )$

Let us evaluate the exponents to get:

${r}^{4} = 4({ {r}^{2} \cos^{2}\theta } - {r}^{2} \sin^{2}\theta)$

Factor the RHS to get:

${r}^{4} = 4{r}^{2} ({ \cos^{2}\theta } - \sin^{2}\theta)$

Divide through by r²

${r}^{2} = 4 ({ \cos^{2}\theta } - \sin^{2}\theta)$

Apply the double angle identity

$\cos^{2}\theta -\sin^{2}\theta= \cos(2 \theta)$

The polar equation then becomes:

${r}^{2} = 4 \cos2\theta$

7 0

3 years ago