1. Consider the curve y=x^2.
a. write down (dy)/(dx)
My answer: 2x
The point P(3,9) lies on the curve y=x^2.
b. Find the gradient of the tangent to the curve at P.
My answer: 2*3=6
c. Find the equation of the normal to the curve at P. Give your answer in the form y=mx+c.
My answer: y=6x-9 but the answer key says y=-1/6x+9.5. Why and how??
2. f(x)=5x^3-3x^5+1 for -1.5<=x<=1.5 and -6<=y<=6.
a. Write down f'(x)
My answer: f'(x)=15x^2-15x^4
b. Find the equation of the tangent to the graph of y=f(x) at (1,3).
My answer:y=x+1 but the answer key says f'(1)=0
Why and how? Isn't asking for EQUATION?
c. Write down the coordinates of the second point where this tangent intersects the graph of y=f(x).
I don't understand what this is asking and am stuck!
3. A small manufacturing company makes and sells x machines each month. The monthly cost C, in dollars, of making x machines is given by
C(x)= 2600+0.4x^2
The monthly income I, in dollars, obtained by selling x machines is given by
I(x)=150x-0.6x^2
P(x) is the monthly profit obtained by selling x machines.
a. Find P(x).
b. Find the number of machines that should be made and sold each month to maximize P(x).
c. Use yours answer to part b to find the selling price of each machine in order to maximize P(x).
1. a) and b) are correct
c) the normal is perpendicular to the tangent
So if the slope of the tangent at (3,9) is 6
then the slope of the normal is -1/6
y = (-1/6)x + b, plug in the point (3,9)
9 = (-1/6)(3) + b
b = 19/2
y = (-1/6)x + 19/2 or y = (-1/6)x + 9.5
2. a) good!
g) if f ' (x) = 15x^2 - 15x^4 , then
f '(1) = 15-15 = 0
so slope = 0, which is a horizontal line
(You have a slope of 1)
A horizontal line through the point is (1,3)
is y = 3 or f(x) = 3
I don't see your answers to #3
for #2c, there is a second point at about (-1.38,3)
see the graph at
http://www.wolframalpha.com/input/?i=solve+5x^3-3x^5%2B1+%3D+3
1. Consider the curve y = x^2.
a. To find (dy)/(dx), we take the derivative of y with respect to x.
dy/dx = 2x
Your answer is correct: (dy)/(dx) = 2x.
b. To find the gradient of the tangent to the curve at point P(3, 9), substitute x = 3 into the derivative:
dy/dx = 2(3) = 6
Your answer is correct: The gradient of the tangent at P is 6.
c. To find the equation of the normal to the curve at P, we need to find the negative reciprocal of the gradient of the tangent. The negative reciprocal of 6 is -1/6.
Using the point-slope form of a line (y - y1) = m(x - x1), we can substitute m = -1/6 and the coordinates of P(3, 9) to find the equation:
y - 9 = (-1/6)(x - 3)
y - 9 = (-1/6)x + 1/2
y = (-1/6)x + 9.5
Therefore, the equation of the normal to the curve at P is y = (-1/6)x + 9.5.
The answer key is correct.
2. Consider the function f(x) = 5x^3 - 3x^5 + 1 for -1.5 <= x <= 1.5 and -6 <= y <= 6.
a. To find f'(x), we take the derivative of f(x) with respect to x.
f'(x) = 3(5)(x^2) - 5(3)(x^4)
f'(x) = 15x^2 - 15x^4
Your answer is correct: f'(x) = 15x^2 - 15x^4.
b. To find the equation of the tangent to the graph of f(x) at point (1, 3), substitute x = 1 into f'(x):
f'(1) = 15(1)^2 - 15(1)^4
f'(1) = 15 - 15
f'(1) = 0
Your answer is incorrect: The gradient of the tangent at (1, 3) is not 1. The derivative f'(x) evaluates to 0 at x = 1, indicating that the tangent is horizontal.
Therefore, the equation of the tangent at (1, 3) is y = 3.
c. The question is asking for the coordinates of the second point where the tangent intersects the graph of y = f(x). Since the tangent is horizontal, it will intersect the graph of y = f(x) at another point where y = 3.
Therefore, the second point of intersection is (x, 3), where -1.5 <= x <= 1.5.
3. Let's consider the monthly cost C(x) and monthly income I(x) equations provided:
C(x) = 2600 + 0.4x^2
I(x) = 150x - 0.6x^2
a. To find the monthly profit P(x), we subtract the cost from the income:
P(x) = I(x) - C(x)
= (150x - 0.6x^2) - (2600 + 0.4x^2)
= 150x - 0.6x^2 - 2600 - 0.4x^2
= -1x^2 + 150x - 2600
The monthly profit P(x) is equal to -1x^2 + 150x - 2600.
b. To maximize the profit P(x), we need to find the value of x that corresponds to the maximum value of P(x). This can be done by finding the x-coordinate of the vertex of the parabola given by P(x).
The x-coordinate of the vertex of a quadratic equation in the form ax^2 + bx + c is given by -b/(2a). In our case, a = -1, b = 150.
x = -b/(2a) = -150/(2*(-1)) = -150/(-2) = 75
Therefore, the number of machines that should be made and sold each month to maximize P(x) is 75.
c. To find the selling price of each machine in order to maximize P(x), we substitute x = 75 into the income equation I(x):
I(x) = 150x - 0.6x^2
= 150(75) - 0.6(75)^2
= 11250 - 0.6(5625)
= 11250 - 3375
= 7875
Therefore, the selling price of each machine needed to maximize P(x) is $7875.
1. Consider the curve y=x^2.
a. To find (dy)/(dx), we need to differentiate the curve with respect to x. In this case, the derivative is found by applying the power rule. Differentiating x^2 gives us 2x. So (dy)/(dx) = 2x.
b. To find the gradient (slope) of the tangent to the curve at point P(3,9), we substitute x=3 into the derivative we found earlier. So the gradient is 2(3) = 6.
c. To find the equation of the normal to the curve at point P, we know that the normal is perpendicular to the tangent. Since the gradient of the tangent is 6, the gradient of the normal is the negative reciprocal, which is -1/6. We can then substitute the coordinates of point P into the equation y=mx+c and solve for c. Using the point (3,9), we have 9=(-1/6)(3)+c. Solving for c, we get c=9.5. Therefore, the equation of the normal is y=(-1/6)x+9.5.
2. Consider the function f(x)=5x^3-3x^5+1 for -1.5<=x<=1.5 and -6<=y<=6.
a. To find f'(x), we need to differentiate the function with respect to x. In this case, we use the power rule to differentiate each term. So, f'(x) = 15x^2 - 15x^4.
b. To find the equation of the tangent to the graph of y=f(x) at the point (1,3), we substitute x=1 into the derivative we found earlier. So f'(1) = 15(1)^2 - 15(1)^4 = 0. Since the derivative is 0 at this point, it means the tangent is horizontal. Therefore, the equation of the tangent is y=3.
c. The question is asking for the coordinates of the second point where this tangent intersects the graph of y=f(x). Since the tangent is horizontal, it means it intersects the graph at multiple points. To find the coordinates, we need to substitute y=3 back into the original function and solve for x. So, we have 3 = 5x^3 - 3x^5 + 1. This is a polynomial equation that can be solved numerically or algebraically to find the values of x and corresponding y.
3. Consider the manufacturing company's monthly cost function C(x) = 2600 + 0.4x^2 and the monthly income function I(x) = 150x - 0.6x^2.
a. To find the monthly profit function P(x), we subtract the monthly cost from the monthly income: P(x) = I(x) - C(x). So, P(x) = (150x - 0.6x^2) - (2600 + 0.4x^2).
b. To maximize the monthly profit P(x), we need to find the value of x that maximizes P(x). One way to do this is by finding the critical points of P(x), which are the points where the derivative P'(x) = 0. We can differentiate P(x) with respect to x, set it equal to 0, and solve for x. Then we can check if the critical points are maxima or minima using the second derivative or by evaluating the points around the critical points. Once we find the x value that maximizes P(x), we can find the corresponding value of P(x) as the maximum monthly profit.
c. To find the selling price of each machine that maximizes the monthly profit P(x), we can divide the monthly profit by the number of machines sold x. So, selling price = P(x)/x. We can substitute the value of x found in part b and calculate the selling price accordingly.