The expression dy/dx = x*(cube root y) gives the slope at any point of the graph of the function f(x) where f(2) = 8.
A. Write the equation of the tangent line to f(x) and any point (2,8).
B. Write an expression for f(x) in terms of x.
C. What is the domain of f(x)?
D. What is the minimum value of f(x)?
Thank you so much!!
y'=x∛y
The tangent line at (2,8) is thus
y-8 = 4(x-2)
y'=x∛y
y^(-1/3) dy = x dx
3/2 y^(2/3) = 1/2 x^2 + c
y^(2/3) = 1/3 x^2 + c
since y(2) = 8,
4 = 4/3 + c
c = 8/3
y^(2/3) = (x^2 + 8)/3
The domain is clearly where x^2+8 >= 0, or all reals.
for minimum value, y'=0, or where x=0 or y=0. Since y is never zero, the minimum is at x=0. Just plug it in.
A. Hmm, let's put on our clown shoes and find the equation of the tangent line! We know that the slope at any point (x, y) on the graph is given by dy/dx, which in this case is x*(cube root y). Since we want to find the equation of the tangent line at (2,8), we can plug those values in. Now, hold on tight as I whip out some math magic!
dy/dx = x*(cube root y)
dy/dx = 2*(cube root 8) (Plug in x = 2 and y = 8)
dy/dx = 2*(2) (The cube root of 8 is 2, bosses!)
So, the slope of the tangent line at (2,8) is 4. Now, we need a point on the line to find the equation. We already have (2,8), so let's get jiggy with it!
Using the point-slope form,
y - y1 = m(x - x1)
y - 8 = 4(x - 2)
And voila! We have the equation of the tangent line to f(x) at (2,8) as y = 4x - 8.
B. Expression time! We wanna find f(x) in terms of x. Brace yourself for some mathematical tomfoolery!
Given dy/dx = x*(cube root y), we can rearrange it to separate variables:
(cube root y) dy = x dx
Now, let's get down to business, put on our thinking caps, and integrate both sides:
∫(cube root y) dy = ∫x dx
Integrating the left side requires some extra circus tricks, as we need to use the u-substitution. But don't worry, I'll be your helpful clown guide!
Let's set u = y^(1/3). Oh, what a lovely letter!
Then, du = (1/3) * y^(-2/3) * dy. I love those fancy symbols!
Substituting back in, we have:
∫u du = ∫x dx
(integral of u with respect to u) = (integral of x with respect to x)
Now, we integrate like there's no tomorrow:
(u^2)/2 + C1 = (x^2)/2 + C2
Keep in mind that C1 and C2 are constants of integration. You know, like your constant need for laughter!
Now, let's substitute back in u:
(y^(1/3))^2/2 + C1 = (x^2)/2 + C2
(y^(2/3))/2 + C1 = (x^2)/2 + C2
C1 - C2 = (x^2)/2 - (y^(2/3))/2
And there you have it, my clown friend! The expression for f(x) in terms of x is:
(x^2)/2 - (y^(2/3))/2 = C1 - C2
C. Let's determine the domain of the function f(x). Remember, clowns always check every path before embarking on their journeys!
The expression dy/dx = x*(cube root y) tells us that the slope at any point (x, y) will exist as long as y^(1/3) is defined. In other words, y must be non-negative for the cube root to be defined. Otherwise, we might end up in some truly wonky clown situations!
So, the domain of f(x) is any x for which the corresponding y-values make the cube root of y a real number, i.e., y ≥ 0.
D. The minimum value of f(x)? Oh, this is where the clownish fun begins!
Since we don't have a specific function for f(x), we can't determine the minimum value directly. We do know that the slope at any point on the graph is given by dy/dx = x*(cube root y).
If the slope dy/dx is always positive or zero, it means that the function is increasing or constant, but it doesn't have a minimum value.
If the slope dy/dx changes from positive to negative at a specific point, then we can say that the function reaches a minimum value at that point. But without further information, it's impossible to determine the minimum value of f(x).
So let's put on our red noses and laugh at the mysteries of life, my clown friend!
I hope my clownish antics have brought a smile to your face and made the world of math a little more entertaining! If you need any more laughter or assistance, please don't hesitate to ask.
A. To find the equation of the tangent line to f(x) at the point (2, 8), we need to use the point-slope form of a line.
The point-slope form is given by:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line and m is the slope of the line. In this case, the point (2, 8) lies on the graph of f(x), so we can substitute x1 = 2 and y1 = 8 into the equation.
We are given that dy/dx = x * (cube root y). To find the slope at x = 2, we substitute x = 2 and y = 8 into the equation:
dy/dx = (2) * (cube root 8) = 2 * 2 = 4.
So, the slope of the tangent line at x = 2 is 4.
Using the point-slope form, we can substitute the values of x1, y1, and m to find the equation of the tangent line:
y - y1 = m(x - x1)
y - 8 = 4(x - 2)
Expanding and simplifying:
y - 8 = 4x - 8
y = 4x
So, the equation of the tangent line to f(x) at the point (2, 8) is y = 4x.
B. To find an expression for f(x), we need to solve the given differential equation. The differential equation is dy/dx = x * (cube root y).
Separating variables, we get:
dy / (y^(1/3)) = x dx
Integrating both sides:
∫(y^(-1/3)) dy = ∫x dx
∫(1/y^(1/3)) dy = ∫x dx
Let's integrate each side:
3 ∫(1/y^(1/3)) dy = ∫x dx
This gives us:
3(3y^(2/3)) = (1/2)x^2 + C
Simplifying:
9y^(2/3) = (1/2)x^2 + C
This is the expression for f(x) in terms of x.
C. The domain of f(x) is the set of all possible values of x for which the expression 9y^(2/3) = (1/2)x^2 + C is defined. Since y should be non-negative, we need to ensure that (1/2)x^2 + C ≥ 0.
The domain of f(x) is the set of all real numbers x such that (1/2)x^2 + C ≥ 0.
D. To find the minimum value of f(x), we need to find the minimum value of the expression (1/2)x^2 + C. Since the coefficient of x^2 is positive, the expression represents a parabola that opens upward. Therefore, there is no minimum value; the function f(x) does not have a minimum value.
A. To find the equation of the tangent line to f(x) at the point (2, 8), we need to find the slope of the tangent line and use the point-slope form of the equation. We are given the expression dy/dx = x*(cube root y), which represents the slope at any point on the graph of f(x).
Step 1: Find the slope at the point (2, 8).
Substitute x = 2 and y = 8 into the given expression:
dy/dx = 2*(cube root 8)
Simplify:
dy/dx = 2*(2)
dy/dx = 4
The slope of the tangent line at the point (2, 8) is 4.
Step 2: Use the point-slope form of the equation to write the equation of the tangent line.
The point-slope form of the equation is:
y - y1 = m(x - x1)
where (x1, y1) is the given point (2, 8) and m is the slope (4).
Substitute the values into the equation:
y - 8 = 4(x - 2)
y - 8 = 4x - 8
y = 4x
So, the equation of the tangent line to f(x) at the point (2, 8) is y = 4x.
B. To find the expression for f(x) in terms of x, we need to solve the given differential equation dy/dx = x*(cube root y).
Step 1: Separate the variables.
Write the equation as:
dy/(x * cube root y) = dx
Step 2: Integrate both sides with respect to their respective variables.
∫(1/y)*dy = ∫x*dx
Step 3: Integrate and simplify.
ln|y| = (1/2)x^2 + C1
Step 4: Solve for y.
Apply the exponential function to both sides to eliminate the natural logarithm:
|y| = e^((1/2)x^2 + C1)
Since |y| represents the absolute value, we can ignore the absolute value sign for simplicity:
y = ±e^((1/2)x^2 + C1)
Let's replace ±e^C1 with a new constant C2:
y = Ce^(1/2)x^2
So, the expression for f(x) in terms of x is f(x) = Ce^(1/2)x^2, where C is a constant.
C. To determine the domain of f(x), we need to consider the values of x that make the expression valid.
In our case, f(x) = Ce^(1/2)x^2, where C is a constant.
The exponential function e^(1/2)x^2 is defined for all real numbers.
Therefore, the domain of f(x) is all real numbers, (-∞, +∞).
D. To find the minimum value of f(x), we need to analyze the concavity of the function f(x).
The function f(x) = Ce^(1/2)x^2 is an exponential function with a positive leading coefficient. This means the graph of f(x) opens upward and has no maximum value.
Since the graph has no maximum, there is no minimum value of f(x).