Suppose f (x) = x^3 - 1 . Find the graph of 5f (x)

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

5 0

Answer:

Graph 1

Step-by-step explanation:

If f(x) =x³- 1

Then 5f (x) 5(x³-1) = 5x³-5

See attached

You might be interested in

Solve this equation pls

Aleksandr-060686 [28]

Answer:

${ \rm{y = \frac{3 \sqrt{x} + 2x}{ {4x}^{2} } }} \\$

• We are gonna use the quotient rule

${ \boxed{ \bf{ \frac{dy}{dx} = \frac{ {\huge{v} } \frac{du}{dx} - { \huge{u}} \frac{dv}{dx} }{ {v}^{2} } }}} \\$

u is (3√x + 2x)
v is 4x²
du/dx is 3/2√x
dv/dx is 8x

• Therefore:

${ \rm{ \frac{dy}{dx} = \frac{(4 {x}^{2})( \frac{3}{2 \sqrt{x} }) - (3 \sqrt{x} + 2x)(8x) }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{(6 {x}^{2})( {x}^{ - \frac{1}{2} } ) - (24 {x}^{ \frac{3}{2} } + 16 {x}^{2} ) }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{ 6 {x}^{ \frac{3}{2} } - {24x}^{ \frac{3}{2} } - {16x}^{2} }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{ - 18 {x}^{ \frac{3}{2} } - 16 {x}^{2} }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = - 2 {x}^{ \frac{3}{2} } (9 + 8 {x}^{ \frac{4}{3} } ) \div 16 {x}^{4} }} \\ \\ { \boxed{ \boxed{ \rm{ \: \: \frac{dy}{dx} = - \frac{8 \sqrt{ {x}^{ \frac{8}{3} }} + 9}{8x {}^{ \frac{8}{3} } } \: \: }}}}$

5 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B2x-24%3D0" id="TexFormula1" title="x^{2} +2x-24=0" alt="x^{2} +2x-24=0" align

denpristay [2]

I used the quadratic formula and got x= 4, -6

8 0

3 years ago

As a first step in solving the system shown, Yumiko multiplies both sides of the equation 2x – 3y = 12 by 6. By what factor shou

GenaCL600 [577]

Answer:

Yumiko should multiply the other equation by 3.

If she adds the two equations she would be left with the variable 'x'.

Step-by-step explanation:

Given the two equations are as follows:

$2x - 3y = 12 \hspace{5mm} \hdots (1)$

$5x + 6y = 18 \hspace{5mm} \hdots (2)$

It is given that she multiplies the first equation by 6. Therefore, (1) becomes

$12x - 18y = 72 \hspace{15mm} \hdots (a)$

Now, note that the sign of the variable 'y' is negative. So, if we make the co-effecient of 'y' equal in both the cases, add them it would result in the elimination of the variable 'y'.

The co-effecient of y in Equation (2) is 6. To make it 18 like it is in Equation (1), we multiply throughout by 3.

Therefore, Equation (2) becomes:

$15x + 18y = 54 \hspace{5mm} \hdots (b)$