#1. A cubic polynomial function f is defined by f(x) = 4x^3 +ax^2 + bx + k
where a, b and k are constants. The function f has a local minimum at x = -1, and the graph of f has a point of inflection at x= -2
a.) Find the values of a and b
#2. Let h be a function defined for all x (not equal to) 0, such that h(4) = -3 and the derivative of h is given by h'(x) = (x^2 - 2) / (x) for all x (not equal to) 0.
a.) Find all values of x for which the graph of h has a horizontal tangent, and determine whether h has a local maximum, a local minimum, or neither at each of these values. Justify your answers.
b.) On what intervals, if any, is the graph of h concave up? Justify
c.) Write an equation for the line tangent to the graph of h at x=4
d.) Does the line tangent to the graph of h at x = 4 lie above or below the graph of h for x > 4 ? Why?
I will do #1 for you.
Show some work or steps that you have done for #2 and#4 and I will evaluate your work.
#1:
f '(x) = 12x^2 + 2ax + b
f ''(x) = 24x + 2a
f '(-1) = 0 = 12 - 2a +b ---> 2a - b = 12
f ''(-2) = 0 = -48 + 2a ----> a= 24
then 2(24) - b = 12
b = 36
a = 24 , b = 36
uytuyt
7
#1.
To find the values of a and b, we need to use the information given about the local minimum and the point of inflection.
First, let's find the derivative of f(x) to determine the critical points:
f'(x) = 12x^2 + 2ax + b
To find the local minimum at x = -1, we set the derivative equal to 0 and solve for x:
12(-1)^2 + 2a(-1) + b = 0
12 - 2a + b = 0
We can use this equation to express b in terms of a:
b = 2a - 12
Next, let's find the second derivative of f(x) to determine the point of inflection:
f''(x) = 24x + 2a
Since the graph of f has a point of inflection at x = -2, we set the second derivative equal to 0 and solve for x:
24(-2) + 2a = 0
-48 + 2a = 0
2a = 48
a = 24
Now that we have a value for a, we can substitute it into the equation for b:
b = 2(24) - 12
b = 48 - 12
b = 36
Therefore, the values of a and b are a = 24 and b = 36.
#2.
a.) To find the values of x for which the graph of h has a horizontal tangent, we need to find where the derivative, h'(x), is equal to zero.
h'(x) = (x^2 - 2) / x
Setting h'(x) equal to zero and solving for x:
(x^2 - 2) / x = 0
x^2 - 2 = 0
x^2 = 2
x = ±√2
So, the graph of h has horizontal tangents at x = √2 and x = -√2.
To determine whether h has a local maximum, a local minimum, or neither at these values, we can analyze the second derivative of h. Let's find h''(x):
h''(x) = (2x * x - (x^2 - 2)) / x^2
h''(x) = (2x^2 - x^2 + 2) / x^2
h''(x) = (x^2 + 2) / x^2
Plugging in x = √2 and x = -√2, we get:
h''(√2) = (√2^2 + 2) / √2^2
h''(√2) = (2 + 2) / 2
h''(√2) = 4 / 2
h''(√2) = 2
h''(-√2) = (-√2^2 + 2) / -√2^2
h''(-√2) = (2 + 2) / 2
h''(-√2) = 4 / 2
h''(-√2) = 2
Since the second derivative is positive (h'' > 0) at both x = √2 and x = -√2, it implies that the graph of h has a local minimum at both of these points.
b.) For the graph of h to be concave up, we need the second derivative, h''(x), to be positive. From our calculations in part a, we found that h''(x) = (x^2 + 2) / x^2.
Since (x^2 + 2) / x^2 is always positive when x ≠ 0, it means that the graph of h is concave up for all x-values except x = 0.
c.) To find the equation for the line tangent to the graph of h at x = 4, we need to find the slope and the y-intercept.
The slope can be found by evaluating h'(4):
h'(4) = (4^2 - 2) / 4
h'(4) = (16 - 2) / 4
h'(4) = 14 / 4
h'(4) = 7 / 2
The y-intercept can be found by evaluating h(4):
h(4) = -3
So, the equation for the tangent line at x = 4 can be written as:
y = (7/2)x - 3
d.) To determine whether the tangent line lies above or below the graph of h for x > 4, we can compare their values at a point greater than 4. Let's choose x = 5:
For the tangent line, y = (7/2)(5) - 3 = 17/2
For h(5), we don't have the value of h for x > 4, so we can't make a comparison.
To find the values of a and b in the cubic polynomial function f(x) = 4x^3 + ax^2 + bx + k, given that it has a local minimum at x = -1 and a point of inflection at x = -2, we can use the properties of the derivative.
a.) Local minimum at x = -1:
To find the value of f'(x) at x = -1, we take the derivative of the function f(x):
f'(x) = 12x^2 + 2ax + b
Evaluate f'(-1) = 12(-1)^2 + 2a(-1) + b
Simplify: -12 + (-2a) + b
We know that at x = -1, f'(-1) = 0 because it has a local minimum. Therefore, we have:
-12 + (-2a) + b = 0
b.) Point of inflection at x = -2:
To find the value of f''(x) at x = -2, we take the second derivative of the function f(x):
f''(x) = 24x + 2a
Evaluate f''(-2) = 24(-2) + 2a
Simplify: -48 + 2a
We know that at x = -2, f''(-2) = 0 because it has a point of inflection. Therefore, we have:
-48 + 2a = 0
By solving these two equations simultaneously, we can find the values of a and b.
Now let's move on to the second question:
Given the function h(x) and its derivative h'(x), we can find various properties and equations related to h(x).
a.) Horizontal tangents and local extremes:
To find the values of x for which the graph of h has horizontal tangents, we need to find where h'(x) = 0.
h'(x) = (x^2 - 2) / x
Set h'(x) = 0:
(x^2 - 2) / x = 0
Solve for x: x^2 - 2 = 0
(x + √2)(x - √2) = 0
This gives two possible values for x: x = √2 and x = -√2
To determine whether h has a local maximum, local minimum, or neither at each of these values, we need to analyze the sign of h'(x) around these points. Since h'(x) = (x^2 - 2) / x, we can make a sign chart:
-∞ -√2 0 √2 +∞
+ - | - +
For x < -√2, h'(x) is positive and increasing.
For -√2 < x < 0, h'(x) is negative and decreasing.
For 0 < x < √2, h'(x) is positive and increasing.
For x > √2, h'(x) is negative and decreasing.
We can conclude that:
At x = -√2, h has a local maximum.
At x = √2, h has a local minimum.
b.) Concavity:
To determine the intervals where the graph of h is concave up, we need to find where h''(x) > 0.
h''(x) = (2x^2 - 4x) / x^2
Simplify: h''(x) = 2 - 4 / x
We want h''(x) > 0:
2 - 4 / x > 0
2 > 4 / x
This implies that x > 0.
Therefore, the graph of h is concave up for x > 0.
c.) Equation of the tangent line at x = 4:
To find the equation of the tangent line at x = 4, we need to find the value of h(4) and h'(4).
Given h(4) = -3, and h'(x) = (x^2 - 2) / x, we can evaluate h'(4) = (4^2 - 2) / 4 = 14 / 4 = 7 / 2.
Using the point-slope form of the equation of a line: y - y1 = m(x - x1),
where (x1, y1) is a point on the line and m is the slope, we have:
y - (-3) = (7/2)(x - 4)
Simplify: y + 3 = (7/2)(x - 4)
This is the equation of the tangent line at x = 4.
d.) Position of tangent line relative to h for x > 4:
To determine whether the tangent line lies above or below the graph of h for x > 4, we need to compare their y-values.
Since the equation of the tangent line at x = 4 is y + 3 = (7/2)(x - 4), we can substitute any value of x > 4 into this equation to find the corresponding y-value on the tangent line.
Now, let's compare the y-values of h and the tangent line:
If the y-value of h is greater than the y-value of the tangent line for x > 4, then the tangent line lies below h.
If the y-value of h is less than the y-value of the tangent line for x > 4, then the tangent line lies above h.
To determine this comparison, we need to evaluate h(x) for x > 4.
However, without specific values for constants or more information, it is not possible to provide a conclusive answer to whether the tangent line lies above or below the graph of h for x > 4.