Math Problem Statement

Is it possible to find lim𝑓(π‘₯)π‘₯β†’0𝑓(π‘₯) and limπ‘₯β†’0𝑔(π‘₯) both do not exist. If 𝑔(π‘₯)while limπ‘₯β†’0possible, give an example.b. Use chain rule to find the derivative of √√π‘₯4 3c. A differentiable function 𝑓 satisfies 𝑓(3) = 5, 𝑓(9) = 7, 𝑓′(3) = 5, 𝑓′(9) = 13. Find an equation for the tangent line to the curve 𝑦 = 𝑓(π‘₯2) at the point (π‘₯, 𝑦) =(3,7).

Solution

a) Finding if lim𝑓(π‘₯) as π‘₯ β†’ 0 and lim𝑔(π‘₯) as π‘₯ β†’ 0 both do not exist:

Yes, it is possible for both limits to not exist. For example, consider two functions:

  • f(x)=sin⁑(1x)f(x) = \sin\left(\frac{1}{x}\right) for xβ‰ 0x \neq 0 and f(0)f(0) is undefined. As xβ†’0x \to 0, the sine function oscillates between -1 and 1 infinitely, so the limit of f(x)f(x) as xβ†’0x \to 0 does not exist.
  • g(x)=1xg(x) = \frac{1}{x} for xβ‰ 0x \neq 0 and g(0)g(0) is undefined. As xβ†’0x \to 0, the function approaches infinity, and thus the limit does not exist either.

Example functions where the limits do not exist:

  • f(x)=sin⁑(1x)f(x) = \sin\left(\frac{1}{x}\right)
  • g(x)=1xg(x) = \frac{1}{x}

b) Using the chain rule to find the derivative of x43\sqrt[3]{x^4}:

To find the derivative of y=x43y = \sqrt[3]{x^4}, first rewrite it as:

y=(x4)1/3=x4/3y = \left( x^4 \right)^{1/3} = x^{4/3}

Now, apply the chain rule:

dydx=ddx(x4/3)\frac{dy}{dx} = \frac{d}{dx} \left( x^{4/3} \right)

Using the power rule, we get:

dydx=43x(4/3)βˆ’1=43x1/3\frac{dy}{dx} = \frac{4}{3}x^{(4/3)-1} = \frac{4}{3}x^{1/3}

So the derivative is:

dydx=43x1/3\frac{dy}{dx} = \frac{4}{3}x^{1/3}

c) Finding the equation for the tangent line to the curve y=f(x2)y = f(x^2) at (x,y)=(3,7)(x, y) = (3, 7):

Given that f(3)=5f(3) = 5, f(9)=7f(9) = 7, fβ€²(3)=5f'(3) = 5, and fβ€²(9)=13f'(9) = 13, we need to find the tangent line to the curve y=f(x2)y = f(x^2) at the point (3,7)(3, 7).

We start by applying the chain rule to differentiate y=f(x2)y = f(x^2). Let u=x2u = x^2, so y=f(u)y = f(u). Then, using the chain rule:

dydx=fβ€²(u)β‹…dudx\frac{dy}{dx} = f'(u) \cdot \frac{du}{dx}

We know u=x2u = x^2, so:

dudx=2x\frac{du}{dx} = 2x

Thus, the derivative of yy is:

dydx=fβ€²(x2)β‹…2x\frac{dy}{dx} = f'(x^2) \cdot 2x

Now, evaluate this derivative at x=3x = 3:

dydx∣x=3=fβ€²(9)β‹…2(3)=13β‹…6=78\frac{dy}{dx}\Big|_{x=3} = f'(9) \cdot 2(3) = 13 \cdot 6 = 78

Now we know the slope of the tangent line at x=3x = 3 is 78. The point of tangency is (3,7)(3, 7).

So, using the point-slope form of the equation of a line:

yβˆ’y1=m(xβˆ’x1)y - y_1 = m(x - x_1)

Substitute m=78m = 78, x1=3x_1 = 3, and y1=7y_1 = 7:

yβˆ’7=78(xβˆ’3)y - 7 = 78(x - 3)

Simplifying:

yβˆ’7=78xβˆ’234y - 7 = 78x - 234

y=78xβˆ’227y = 78x - 227

So, the equation of the tangent line is:

y=78xβˆ’227y = 78x - 227


Would you like further details on any of these solutions? Or do you have any other questions?


Here are some additional questions to explore:

  1. How do limits behave for oscillatory functions like sin⁑(1x)\sin\left( \frac{1}{x} \right)?
  2. What are other examples of functions that do not have limits as x→0x \to 0?
  3. Can you apply the chain rule to more complex compositions, such as f(g(x))f(g(x)) where both functions are non-polynomial?
  4. How can you use the power rule for derivatives with fractional exponents?
  5. What is the geometric interpretation of the derivative of a function at a point?

Tip: When working with the chain rule, remember to treat composite functions step-by-step, differentiating the outer function first and then multiplying by the derivative of the inner function.

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Math Problem Analysis

Mathematical Concepts

Limits
Chain Rule
Derivative
Tangent Line
Differentiation

Formulas

lim x→0 sin(1/x) does not exist
lim x→0 1/x does not exist
Chain Rule: d/dx[f(g(x))] = f'(g(x)) * g'(x)
Power Rule: d/dx x^n = n * x^(n-1)
Point-slope form of a line: y - y1 = m(x - x1)

Theorems

Limit of a function
Chain rule of differentiation
Derivative of composite functions
Equation of the tangent line

Suitable Grade Level

College Level (Calculus)