Question

Attached below are my calculus questions. If you are able to help, thank you...!

Answer 1

a.
`\( f'(-10) = -1200 \).`

b.
`\( f^(-1)(y) = \boxed{-\frac{\sqrt[3]{2(y + 3)}}{2}} \).`

c.
`\( \left(f^(-1)\right)'(3997) = -(1)/(1200) \).`

How did we get the value?

a. Find
`\( (df)/(dx) \) at \( x = -10 \)`:

Given the function
`\( y = f(x) = -4x^3 - 3 \)`, we can find the derivative
`\( (df)/(dx) \)` by differentiating each term:

`\[ (df)/(dx) = -12x^2 \]`

Now, evaluate
`\( (df)/(dx) \) at \( x = -10 \):`

`\[ (df)/(dx)(-10) = -12(-10)^2 \\= -12(100) \\= -1200 \]`

So,
`\( f'(-10) = -1200 \).`

b. Find a formula for
`\( x = f^(-1)(y) \)`:

To find the inverse function
`\( f^(-1)(y) \)`, interchange
`\( x \)` and
`\( y \)` and solve for
`\( y \)`:

`\[ y = -4x^3 - 3 \]`

Interchange
`\( x \)` and
`\( y \)`:

`\[ x = -4y^3 - 3 \]`

Now, solve for
`\( y \)`:

`\[ 4y^3 = -x - 3 \]`

`\[ y^3 = -(x + 3)/(4) \]`

`\[ y = \sqrt[3]{-(x + 3)/(4)} \]`

So,
`\( f^(-1)(y) = \boxed{-\frac{\sqrt[3]{2(y + 3)}}{2}} \).`

c. Find
`\( (df^(-1))/(dy) \)`at
`\( y = f(-10) \)`:

Evaluate
`\( f(-10) \)` using the original function:

`\[ f(-10) = -4(-10)^3 - 3 \\= -4(-1000) - 3 \\= 4000 - 3 \\= 3997 \]`

Now, find
`\( (df^(-1))/(dy) \)` at
`\( y = 3997 \)`:

`\[ (df^(-1))/(dy)(3997) = (1)/(f'\left(f^(-1)(3997)\right)) \]`

We know
`\( f^(-1)(3997) = -10 \)`, and we previously found
`\( f'(-10) = -1200 \)`.

`\[ (df^(-1))/(dy)(3997) = (1)/(-1200) \\= -(1)/(1200) \]`

So,
`\( \left(f^(-1)\right)'(3997) = -(1)/(1200) \).`

Answer 2

`\textsf{a)} \quad f'(-10)=-1200`

`\textsf{b)}\quad f^(-1)(y)=-\frac{\sqrt[3]{2(y+3)}}{2}`

`\textsf{c)} \quad \left(f^(-1)\right)'(f(-10))=-(1)/(1200)`

Explanation:

Part a

Given equation:

`y=f(x)=-4x^3-3`

Differentiate f(x) using the power rule and the constant rule:

`f'(x)=3 \cdot -4x^(3-1)-0`

`f'(x)=-12x^2`

To find f'(-10), substitute x = -10 into f'(x):

`f'(-10)=-12(-10)^2`

`f'(-10)=-12(100)`

`f'(-10)=-1200`

Therefore, the value of f'(-10) is:

`\large\boxed{\boxed{f'(-10)=-1200}}`

`\hrulefill`

Part b

To find the formula for x = f⁻¹(y), rearrange the equation to isolate x then replace x with f⁻¹(y):

`y=-4x^3-3`

`y+3=-4x^3`

`(y+3)/(4)=-x^3`

`-x=\sqrt[3]{(y+3)/(4)}`

`x=-\sqrt[3]{(y+3)/(4)}`

Replace x with f⁻¹(y):

`f^(-1)(y)=-\sqrt[3]{(y+3)/(4)}`

To eliminate the radical denominator, multiply by ∛2 / ∛2:

`f^(-1)(y)=-\frac{\sqrt[3]{y+3}}{\sqrt[3]{4}}\cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}}`

`f^(-1)(y)=-\frac{\sqrt[3]{2(y+3)}}{\sqrt[3]{4\cdot2}}`

`f^(-1)(y)=-\frac{\sqrt[3]{2(y+3)}}{\sqrt[3]{8}}`

`f^(-1)(y)=-\frac{\sqrt[3]{2(y+3)}}{\sqrt[3]{2^3}}`

`f^(-1)(y)=-\frac{\sqrt[3]{2(y+3)}}{2}`

Therefore, the formula for x = f⁻¹(y) is:

`\large\boxed{\boxed{f^(-1)(y)=-\frac{\sqrt[3]{2(y+3)}}{2}}}`

`\hrulefill`

Part c

To find df⁻¹/dy at y = f(-10), we first need to differentiate f⁻¹(y) with respect to y. To do this, we can use the chain rule.

`\boxed{\begin{array}{c}\underline{\text{Chain Rule for Differentiation}}\\\\\text{If\;\;$y=f(u)$\;\;and\;\;$u=g(x)$\;\;then:}\\\\\frac{\text{d}x}{\text{d}y}=\frac{\text{d}x}{\text{d}u}*\frac{\text{d}u}{\text{d}y}\end{array}}`

`\textsf{Let}\;\;x=-(1)/(2)\sqrt[3]{u}\;\;\textsf{where}\;\;u=2y+6`

Differentiate the two parts separately:

`\begin{aligned}x=-(1)/(2)u^(\frac13) \implies \frac{\text{d}x}{\text{d}u}&=(1)/(3)\cdot -(1)/(2)u^(\frac13-1)\\\\\frac{\text{d}x}{\text{d}u}&=-(1)/(6)u^(-\frac23)\\\\\frac{\text{d}x}{\text{d}u}&=-(1)/(6u^(\frac23))\end{aligned}`

`u=2y+6\implies \frac{\text{d}u}{\text{d}y}=2\\`

Now, put everything back into the chain rule formula:

`\frac{\text{d}x}{\text{d}y}=-(1)/(6u^(\frac23))*2`

`\frac{\text{d}x}{\text{d}y}=-(1)/(3u^(\frac23))`

Substitute back in u = 2y + 6:

`\frac{\text{d}x}{\text{d}y}=-(1)/(3(2y+6)^(\frac23))`

Therefore:

`\frac{\text{d}f^(-1)}{\text{d}y}=(f^(-1))'(y)=-(1)/(3(2y+6)^(\frac23))`

Find the value of y when x = -10:

`\begin{aligned}y=f(-10)&=-4(-10)^3-3\\\\ &=-4(-1000)-3\\\\ &=4000-3\\\\ &=3997\end{aligned}`

Therefore, to find df⁻¹/dy at y = f(-10), substitute y = 3997 into (f⁻¹)'(y):

`\begin{aligned}\left(f^(-1)\right)'(f(-10))=(f^(-1))(3997)&=-(1)/(3(2(3997)+6)^(\frac23))\\\\&=-(1)/(3(8000)^(\frac23))\\\\&=-(1)/(3(400))\\\\&=-(1)/(1200)\end{aligned}`

Therefore, the value of df⁻¹/dy at y = f(-10) is:

`\large\boxed{\boxed{\left(f^(-1)\right)'(f(-10))=-(1)/(1200)}}`