All categories

2 years ago

If f(x)=3 x -1 and g(x)=0.03x^3-x+1 an approximate solution for the equation f(x)=g(x) is

Mathematics

1 answer:

lubasha [3.4K]2 years ago

3 0

Answer:

x ≈ {-11.789, +0.501, 11.288}

Step-by-step explanation:

The cubic g(x) - f(x) = 0 has three real solutions. It can be rewritten as ...

$g(x)-f(x)=0\\\\0.03x^3-x+1-(3x-1)=0\\\\0.03x^3-4x+2=0$

Since the solutions are irrational, they are best found using a spreadsheet or graphing calculator. My favorite graphing calculator shows the approximate solutions below.

_____

Comment on the problem statement

Your expression for f(x) is ambiguous in that many Brainly questions have the exponentiation indicator replaced by a blank: 3 x -1 often means 3^x -1 and sometimes means 3^(x-1). We have taken the expression at face value and have assumed it is a linear expression. If otherwise, the problem is basically worked the same way: write a function h(x) = g(x) - f(x) and look for solutions to h(x) = 0. Graphing can be useful.

You might be interested in

B Identify the coefficient and the number of terms in the

Elenna [48]

It’s D because the number that have an (x) is a COEFFICIENT and I hope it helps :)

8 0

2 years ago

Please help. Screenshot attachedn

NARA [144]

Answer:

x = -0.17 and x = -5.83

Step-by-step explanation:

We are asked to solve the quadratic equation

$x^2+6\,x+1=0$

We use the quadratic formula using a = 1, b = 6 and c = 1

for a general quadratic equation of the form: $ax^2+bx+c=0$

Then, the solutions are given by;

$x=\frac{-6+/-\sqrt{36-4\,(1)\,(1)} }{2} =\frac{-6+/-\sqrt{32} }{2}$

which produces the two following answers (rounded to two decimals):

x = -0.17 and x = -5.83

8 0

3 years ago

Which answer best describes the complex zeros of the polynomial function?

marysya [2.9K]

Answer-

D. The function has one real zero and two nonreal zeros. The graph of the function intersects the x-axis at exactly one location.

Solution-

The given polynomial is,

$f(x)=x^3-x^2+6x-6$

The zeros of the polynomials are,

$\Rightarrow f(x)=0$

$\Rightarrow x^3-x^2+6x-6=0$

$\Rightarrow x^2(x-1)+6(x-1)=0$

$\Rightarrow (x^2+6)(x-1)=0$

$\Rightarrow (x^2+6)=0,(x-1)=0$

$\Rightarrow x=\sqrt{-6},\ x=1$

$\Rightarrow x=1,\ \pm \sqrt6i$

Therefore, this function has only one real zero i.e 1 and two nonreal zeros i.e ±√6i . The graph of the function intersects the x-axis at exactly one location i.e at x = 1