How do you sketch the graph #f(x)=x^3+1#?

Answer 1

see explanation

write as:#" " y=x^3+1#

At #x=0; y=1 larr " y-intercept"#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#(dy)/(dx)=3x^2 => "local max/min at "x=0 -> (0,1)#

Only one #x# value for local max/min so the graph gradient goes from positive to 0 to positive again (no negative gradient)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

At #y=0 => x^3=-1 larr" x-intercept"#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For 2 more points set #x=+-2#

At #x=-2: x^3+1=-7-> (-2,-7)#

At #x=2: x^3+1 = +9->(+2,+9)#

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Answer 2

Scarlett Allison

To sketch the graph of $f(x) = x^3 + 1$ , follow these steps:

Identify key points:
- Y-intercept: $f(0) = 0^3 + 1 = 1$
- X-intercepts: $x^3 + 1 = 0$ has no real solutions.
Determine behavior at extreme values:
- As $x$ approaches negative infinity, $f(x)$ approaches negative infinity.
- As $x$ approaches positive infinity, $f(x)$ approaches positive infinity.
Determine the direction of the graph:
- Since the leading term $x^3$ is positive, the graph rises to the right and falls to the left.
Sketch the graph using the identified points and behavior:
- Draw a curve rising to the right, passing through the point (0, 1), and falling to the left without crossing the x-axis.

The resulting graph resembles a "S" shape, with the left side falling and the right side rising, passing through the point (0, 1).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.