5. Let for and f(x)=12X^2 for x>=0 and f(x)>=0
a. The line tangent to the graph of f at the point (k, f(k)) intercepts the x-axis at x = 4. What is the value of k?
b. An isosceles triangle whose base is the interval from (0, 0) to (c, 0) has its vertex on the graph of f. For what value of c does the triangle have a maximum area? Justify your answer.
dy/dx = 24x
at (k, 12k^2)
dy/dx = 24k
equation of tangent:
y - 12k^2 = 24k(x-k)
but the point (4,0) is on this line
0-12k^2 = 24k(4-k)
-12k^2 = 96k - 24k^2
12k^2 - 96k = 0
k(k-8) = 0
k = 0 or k = 8
at k=0, the tangent would be the x-axis itself, which does pass through the point (4,0)
More than likely they want k=8
check:
if k=8 , the point on the curve is (8, 768)
slope of tangent is 96
tangent equation at (8,768) is
y - 768 = 96(x-8)
for x-intercept, let y = 0
-768 = 96x - 768
0 = 96x
x = 0
If the base is form (0,0) to (c,0) and the vertex of the isosceles triangle lies on y = 12x^2, then the altitude must right-bisect the base
So if we let the vertex be (a,12a^2) then the base will be on (0,0 and (2a,0)
Area = (1/2)(2a)(12a^2) = 12a^3
As a gets larger, 12a^3 gets larger and there is no maximum area.
a. To find the value of k, we need to find the equation of the tangent line at the point (k, f(k)) and then find the x-intercept of that line.
The formula for finding the equation of a tangent line at a point (a, f(a)) on a curve is y - f(a) = f'(a)(x - a), where f'(a) is the derivative of f(x) evaluated at x = a.
In this case, f(x) = 12x^2, so f'(x) = 24x.
Let's find the derivative at x = k:
f'(k) = 24k.
The equation of the tangent line is then:
y - f(k) = f'(k)(x - k).
We know that the tangent line intercepts the x-axis at x = 4, so we can substitute these values into the equation:
0 - f(k) = f'(k)(4 - k).
Since f(x) >= 0, this means f(k) >= 0. Therefore, we can simplify the equation to:
0 = f'(k)(4 - k).
Substituting f'(k) = 24k, we get:
0 = 24k(4 - k).
Expanding this equation:
0 = 96k - 24k^2.
Rearranging terms:
24k^2 - 96k = 0.
Factoring out common factor:
24k(k - 4) = 0.
Setting each factor equal to 0:
k = 0 or k - 4 = 0.
Solving for k, we find:
k = 0 or k = 4.
Therefore, the possible values of k are 0 or 4.
b. The area of an isosceles triangle with base b and height h is given by the formula A = (1/2)bh. In this case, the base is the interval from (0, 0) to (c, 0), which has length b = c - 0 = c.
To find the maximum area of the triangle, we need to find the value of c that maximizes the area. We know that the vertex of the triangle lies on the graph of f(x) = 12x^2.
The height of the triangle is given by the value of f(c). Therefore, the formula for the area becomes:
A = (1/2)(c)(f(c)).
Substituting f(c) = 12c^2, we get:
A = (1/2)(c)(12c^2) = 6c^3.
To find the value of c that maximizes the area, we can take the derivative of A with respect to c and set it equal to 0, since the maximum or minimum occurs when the derivative is equal to 0.
Let's find the derivative of A with respect to c:
dA/dc = 18c^2.
Setting dA/dc = 0, we get:
18c^2 = 0.
Since c is a length, it cannot be zero. Therefore, there is no maximum area for the triangle in this scenario.
Justification: The derivative of the area function is always positive for c > 0, indicating that the area is increasing as c increases. Therefore, there is no maximum area for the triangle, and it can be infinitely large as c increases.
a. To find the value of k, we need to find the slope of the tangent line at the point (k, f(k)). The slope of the tangent line is equal to the derivative of the function f(x) evaluated at x = k.
Given that f(x) = 12x^2, we can take the derivative of f(x) with respect to x:
f'(x) = d/dx (12x^2)
= 24x
Now, we can evaluate the derivative at x = k:
f'(k) = 24k
Since the question states that the tangent line intercepts the x-axis at x = 4, this means the y-coordinate of the point of tangency is 0. By substituting x = 4 into the equation of the tangent line, we can find the y-coordinate:
f(4) = 12(4)^2 = 192
So, the point (k, f(k)) is (4, 192).
To determine the value of k, we can set up the equation for the equation of the tangent line:
(y - f(k)) = m(x - k)
Substituting the known values, we get:
(y - 192) = 24k(x - 4)
Since the tangent line intercepts the x-axis at x = 4, we know that y = 0:
(0 - 192) = 24k(4 - 4)
-192 = 0
This equation has no solution, which means there is no value of k that satisfies the given conditions.
b. The area of an isosceles triangle with base length b and height h is given by the formula:
Area = (1/2) * b * h
In this case, the base length is the interval from (0, 0) to (c, 0), which has a length of c.
The height of the triangle is the y-coordinate of the point on the graph of f(x) that is directly above the midpoint of the base.
To find the maximum area, we need to find the value of c that maximizes the height of the triangle. Since f(x) = 12x^2, the height is given by f(c/2).
To find the maximum value of f(c/2), we can take the derivative of f(x) with respect to x:
f'(x) = d/dx (12x^2)
= 24x
Now, we can evaluate the derivative at x = c/2:
f'(c/2) = 24(c/2)
= 12c
To find the maximum value of f(c/2), we set the derivative equal to zero:
12c = 0
This equation implies that c = 0, which means the base length would be 0 as well, resulting in a degenerate triangle.
Therefore, there is no value of c that will result in a triangle with a maximum area on the graph of f(x) = 12x^2.