Ask question

Ask question

All categories

3 years ago

6

F(x)=3x+4 and g(x) 4x^(2)-x Find f(-5) and g(-2)

1 answer:

katovenus [111]3 years ago

7 0

F(-5)=3(-5)+4=-11
g(-2)=4(-2)^2+2=18

You might be interested in

Someone help -2/5 x 1/4 x 1/5

kari74 [83]

Answer:

Lol i got u homie!

-0.02

(Sorry if that wasn't a fraction but hopefully this helps)

Step-by-step explanation:

7 0

2 years ago

How many days are there from march 15 to september 15 inclusive?

Natasha2012 [34]

March 15 - March 31 = 17 days

April = 30days

May = 31 days

June = 30 days

July = 31 days

August = 31 days

Sept 1 - Sept 15 = 15 days

total = 185 days

6 0

3 years ago

I am having trouble with this relative minimum of this equation.<br>

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$

$f'(x)=12x^2+26x-12$

$f''(x)=24x+26$

So if f''(a) is positive then we have a minimum at x=a.

If f''(a) is negative then we have a maximum at x=a.

Rounding to nearest tenths here: x=-2.6 and x=.4

Let's see what f'' gives us at both of these x's.

$24(-2.6)+25$

$-37.5$

So we have a maximum at x=-2.6.

$24(.4)+25$

$9.6+25$

$34.6$

So we have a minimum at x=.4.

Now let's find the corresponding y-value for our relative minimum point since that would complete your question.

We are going to use the equation that relates x and y.

I'm going to use 0.3909838 instead of .4 just so we can be closer to the correct y value.

$y=4(0.3909838)^3+13(0.3909838)^2-12(0.3909838)-56$

I'm shoving this into a calculator:

$y=-58.4654411$

So the approximate relative minimum is (0.4,-58.5).

If you graph $y=4x^3+13x^2-12x-56$ you should see the graph taking a dip at this point.

3 0

3 years ago

In the equation (m^5/6) (m^1/6)^7 = m^x , what is the value of x that makes the equation true?

vlabodo [156]

${m}^{ \frac{5}{6} } \times (m {}^{ \frac{1}{6} } ) {}^{7 } = m {}^{x}$

$m {}^{ \frac{5}{6} } \times m {}^{ \frac{7}{6} } = m {}^{x}$

${m}^{ \frac{5 \times 7}{6 \times 6} } = {m}^{x}$

$\frac{35}{36} = x$

6 0

2 years ago

209.106 in expanded form

Juliette [100K]

It would be 200+9+0.1+0.006.

4 0

3 years ago

Read 2 more answers