# How do u graph this function f(x)=x^3-4? and Approximate to the nearest tenth, the real root of the equation f(x)=x^3-4=0

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To graph a function, you have to find some important points which belongs to the graph, to check if the function is increasing or decreasing, to verify if the function is continuous, to see if the function is convex or concave.

Let's see if the function is continuous:

lim f(x) = lim (x^3-4) = limx^3-lim4 = inf, if x->inf

lim f(x) = -inf, if x->-inf

The function is continuous.

Let's see if the function is increasing or decreasing.

For this rason, we'll calculate it's first derivative:

f'(x) = 3x^2

Because f'(x)>0, no matter if x>0 or x<0, that means that f(x) is an increasing function.

For x=0, the function has an inflection point.

To check the aspect of the function, we have to calculate the second derivative:

f"(x) = 6x

For x<0. f"(x)<0, so f(x) is concave

For x>0, f"(x)>0, so f(x) is convex.

Now, we have to give values for x and we'll find values for y, so that to draw the graph of f(x), which passes through these points.

To graph f(x) = x^3-4.

Solution:

Let y = x^3-4.

First find the coordinayes of some of the points satisfying the y =x^3-4, as below:

Let x =0 , then y = 0^3-4 = -4.

x=1 , y = 1^3-4 =-3

f(1) = -3 and f(2) = 4.

x1 = 1 + {f(0)-f(1)}/f(2)-f(1) = 1+ 3/7 = 1.43.

f(1.43) = -1.076

f(2) = 4.

Therefore x2 = 1.43 +{ 0-f(1.43)/(f(2)-f(4.3)} (2-4.3)

= 1.43 +1.076/(4+1.076)}(2-1.43)

= 1.43+0.12 = 1.55 = 1.6 to the 1st decimal place

x=2 and y = 2^3 -4 = 4.

x =3 , y =3^3-4 = 23

Like that we go any desired number of pairs of coordinates

Plot (0,-4), (1,-3), (2,4) and (3,23 ) in the graph.

We see that the real solution for f(x) lies between x=1 and x=2 as f(1) = -3 and f(2) = 4 and f(x) = o for some value of x i (1,2).