Determine $$ f^{-1}(x) $$ from first principles if $$ f(x)=3 x^{2} $$ John determines $$ g^{\prime}(a) $$, the derivative of a certain function $$ g $$ at $$ x=a $$, and arrives at the answer. $$ \lim _{x \rightarrow 0} \frac{\sqrt{4+h}-2}{h} $$ Write down the equation of $$ g $$ and the value of $$ a $$. Determine $$ \frac{d y}{d x} $$ if $$ y=\sqrt{x^{3}}-\frac{5}{x^{3}} $$ $$ g(x)=-8 x+20 $$ is a tangent to $$ \quad f(x)=x^{3}+a x^{2}+b x+18 $$ at $$ x=1 $$. Calculate the values of $$ a $$ and $$ b $$.

Question

UpStudy AI · Accepted Answer

1. **Finding $$ f^{-1}(x) $$ for $$ f(x)=3x^2 $$:**

To find the inverse we start from
   $$
   y=3x^2.
   $$
   Assuming the domain is restricted to $$ x\ge0 $$ (so that $$ f $$ is one‐to‐one), we solve for $$ x $$:
   $$
   x^2=\frac{y}{3} \quad\Longrightarrow\quad x=\sqrt{\frac{y}{3}}.
   $$
   Interchanging the roles of $$ x $$ and $$ y $$, the inverse function is
   $$
   f^{-1}(x)=\sqrt{\frac{x}{3}}.
   $$

2. **Determining $$ g(x) $$ and $$ a $$ from**
   $$
   \lim_{h\to 0}\frac{\sqrt{4+h}-2}{h}.
   $$
   Recall the definition of the derivative:
   $$
   g'(a)=\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}.
   $$
   Comparing with the given limit, we identify
   $$
   g(a)=2 \quad \text{and} \quad g(a+h)=\sqrt{4+h}.
   $$
   Therefore, it is natural to take
   $$
   g(x)=\sqrt{x}
   $$
   and hence
   $$
   g(4)=\sqrt{4}=2.
   $$
   Thus, $$ a=4 $$.

3. **Calculating $$\frac{dy}{dx}$$ for $$ y=\sqrt{x^3}-\frac{5}{x^3} $$:**

First, rewrite the expression using exponents:
   $$
   y=x^{\frac{3}{2}}-5x^{-3}.
   $$
   Differentiate term by term using the power rule:
   $$
   \frac{d}{dx}\left(x^{\frac{3}{2}}\right)=\frac{3}{2}x^{\frac{1}{2}}, \quad
   \frac{d}{dx}\left(-5x^{-3}\right) = -5(-3)x^{-4}=15x^{-4}.
   $$
   Therefore, the derivative is
   $$
   \frac{dy}{dx}=\frac{3}{2}x^{\frac{1}{2}}+15x^{-4}.
   $$

4. **Finding $$ a $$ and $$ b $$ such that $$ g(x)=-8x+20 $$ is tangent to $$ f(x)=x^3+ax^2+bx+18 $$ at $$ x=1 $$:**

Since $$ g(x) $$ is tangent to $$ f(x) $$ at $$ x=1 $$, the following two conditions must hold:

- **Function values are equal at $$ x=1 $$:**
     $$
     f(1)=g(1).
     $$
     Compute:
     $$
     f(1)=1^3+a\cdot1^2+b\cdot1+18=1+a+b+18=a+b+19,
     $$
     $$
     g(1)=-8(1)+20=12.
     $$
     Thus, we have
     $$
     a+b+19=12 \quad\Longrightarrow\quad a+b=-7. \quad \text{(Equation 1)}
     $$

- **Derivatives are equal at $$ x=1 $$:**
     First, compute $$ f'(x) $$:
     $$
     f'(x)=3x^2+2ax+b.
     $$
     Evaluating at $$ x=1 $$:
     $$
     f'(1)=3(1)^2+2a(1)+b=3+2a+b.
     $$
     The derivative of the tangent line $$ g(x)=-8x+20 $$ is
     $$
     g'(x)=-8.
     $$
     Therefore,
     $$
     3+2a+b=-8 \quad\Longrightarrow\quad 2a+b=-11. \quad \text{(Equation 2)}
     $$

Now, subtract Equation 1 from Equation 2:
   $$
   (2a+b) - (a+b)= -11 - (-7),
   $$
   $$
   a= -4.
   $$
   Substitute $$ a=-4 $$ into Equation 1:
   $$
   -4+b=-7 \quad\Longrightarrow\quad b=-3.
   $$

Hence, the values are $$ a=-4 $$ and $$ b=-3 $$.
1. The inverse function is $$ f^{-1}(x) = \sqrt{\frac{x}{3}} $$. 2. The function $$ g(x) = \sqrt{x} $$ and the value of $$ a $$ is 4. 3. The derivative $$ \frac{dy}{dx} = \frac{3}{2}x^{\frac{1}{2}} + 15x^{-4} $$. 4. The values are $$ a = -4 $$ and $$ b = -3 $$.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Calculus Questions