-x^4-3x^3+5x+6
Where does this function concave up and down?
Thanks for your help!!!
Is it
y = -x^4 - 3x^2 + 5x + 6 ?
if so, take the second derivative.
if y '' is positive , the graph of the function is concave upwards
if y '' is negative , the graph of the function is concave downwards
no its -x^4-3x^3+5x+6
A function has to be written as an equation.
You just have an expression.
Is this not Calculus?
its calculus its a webwork prob for my class
Did you get
-12x^2 - 18x as your second derivative ?
critical values for this are
x = 0 and x = -3/2
( I set these equal to zero and solved)
You know the graph will be 'W"shaped downwards because of the -x^4.
consider the value of -12x^2 - 18x for x < -3/2
(try x = -5)
The expression will be negative, so
concave downwards for x < -3/2
consider the value of -12x^2 - 18x for x between -3/2 and 0
(try x = -1)
The expression will be -12(1) - 18(-1) which is positve,
so concave upwards for -3/2 < x < 0
I will let you decide what happens for x > 0
To determine where the given function is concave up or concave down, you need to find the second derivative of the function. Let's follow the steps:
Step 1: Start with the given function: f(x) = -x^4 - 3x^3 + 5x + 6.
Step 2: Find the first derivative by differentiating the function: f'(x) = -4x^3 - 9x^2 + 5.
Step 3: Now, differentiate the first derivative to find the second derivative: f''(x) = -12x^2 - 18x.
Step 4: Set the second derivative equal to zero and solve for x to find the inflection points, if any.
-12x^2 - 18x = 0
Factor out -6x: -6x(2x + 3) = 0
Set each factor equal to zero:
-6x = 0 --> x = 0
2x + 3 = 0 --> 2x = -3 --> x = -3/2
So, the possible inflection points are x = 0 and x = -3/2.
Step 5: Create a number line with the critical points (-3/2 and 0) and choose test points within each interval to determine the concavity.
Test a point to the left of -3/2, for example, x = -2:
substitute x = -2 into the second derivative: f''(-2) = -12(-2)^2 - 18(-2) = -24 + 36 = 12
Since f''(-2) = 12 is positive, the function is concave up to the left of x = -3/2.
Test a point between -3/2 and 0, for example, x = -1:
substitute x = -1 into the second derivative: f''(-1) = -12(-1)^2 - 18(-1) = -12 - (-18) = 6
Since f''(-1) = 6 is positive, the function is still concave up in the interval (-3/2,0).
Test a point to the right of zero, for example, x = 1:
substitute x = 1 into the second derivative: f''(1) = -12(1)^2 - 18(1) = -12 - 18 = -30
Since f''(1) = -30 is negative, the function is concave down to the right of x = 0.
So, to summarize:
- The function is concave up in the interval (-∞, -3/2) and the interval (-3/2, 0).
- The function is concave down in the interval (0, ∞).
These intervals indicate where the graph of the function is curving upwards (concave up) and where it is curving downwards (concave down).