for each function $g(x)$, its inverse $g^{-1}(x)=f(x)$. evaluate the given derivative.\n13…

for each function $g(x)$, its inverse $g^{-1}(x)=f(x)$. evaluate the given derivative.\n13. $g(x)=cos(x)+3x^{2}$, $g(\frac{pi}{2})=\frac{3pi}{4}$. find $f(\frac{3pi}{4})$\n14. $g(x)=2x^{3}-x^{2}-5x$, $g(-2)=-10$. find $f(-10)$\n15. $g(x)=sqrt{8 - 2x}$. find $f(4)$?\n16. $g(x)=x^{3}-7$. find $f(20)$?\n17. $g(x)=\frac{5}{x + 3}$. find $f(\frac{1}{2})$?

for each function $g(x)$, its inverse $g^{-1}(x)=f(x)$. evaluate the given derivative.\n13. $g(x)=cos(x)+3x^{2}$, $g(\frac{pi}{2})=\frac{3pi}{4}$. find $f(\frac{3pi}{4})$\n14. $g(x)=2x^{3}-x^{2}-5x$, $g(-2)=-10$. find $f(-10)$\n15. $g(x)=sqrt{8 - 2x}$. find $f(4)$?\n16. $g(x)=x^{3}-7$. find $f(20)$?\n17. $g(x)=\frac{5}{x + 3}$. find $f(\frac{1}{2})$?

Answer

Explanation:

Step1: Recall the formula for the derivative of an inverse function

If $y = f(x)$ is the inverse of $g(x)$, then $f^{\prime}(y)=\frac{1}{g^{\prime}(x)}$ where $y = g(x)$.

Step2: For problem 13

First, find the derivative of $g(x)=\cos(x)+3x^{2}$. Using the sum - rule and basic derivative formulas, $g^{\prime}(x)=-\sin(x) + 6x$. Given $g(\frac{\pi}{2})=\frac{3\pi}{4}$, by the formula for the derivative of the inverse function $f^{\prime}(\frac{3\pi}{4})=\frac{1}{g^{\prime}(\frac{\pi}{2})}$. Substitute $x = \frac{\pi}{2}$ into $g^{\prime}(x)$: $g^{\prime}(\frac{\pi}{2})=-\sin(\frac{\pi}{2})+6\times\frac{\pi}{2}=- 1 + 3\pi$. So $f^{\prime}(\frac{3\pi}{4})=\frac{1}{-1 + 3\pi}$.

Step3: For problem 14

Find the derivative of $g(x)=2x^{3}-x^{2}-5x$. Using the power - rule, $g^{\prime}(x)=6x^{2}-2x - 5$. Given $g(-2)=-10$, by the formula for the derivative of the inverse function $f^{\prime}(-10)=\frac{1}{g^{\prime}(-2)}$. Substitute $x=-2$ into $g^{\prime}(x)$: $g^{\prime}(-2)=6\times(-2)^{2}-2\times(-2)-5=24 + 4-5=23$. So $f^{\prime}(-10)=\frac{1}{23}$.

Step4: For problem 15

Find the derivative of $g(x)=\sqrt{8 - 2x}=(8 - 2x)^{\frac{1}{2}}$. Using the chain - rule, $g^{\prime}(x)=\frac{1}{2}(8 - 2x)^{-\frac{1}{2}}\times(-2)=\frac{-1}{\sqrt{8 - 2x}}$. We want to find $f^{\prime}(4)$. First, find $x$ such that $g(x)=4$. Set $\sqrt{8 - 2x}=4$, then $8 - 2x = 16$, $x=-4$. So $f^{\prime}(4)=\frac{1}{g^{\prime}(-4)}$. Substitute $x = - 4$ into $g^{\prime}(x)$: $g^{\prime}(-4)=\frac{-1}{\sqrt{8-2\times(-4)}}=\frac{-1}{\sqrt{16}}=-\frac{1}{4}$. So $f^{\prime}(4)=-4$.

Step5: For problem 16

Find the derivative of $g(x)=x^{3}-7$. Using the power - rule, $g^{\prime}(x)=3x^{2}$. We want to find $f^{\prime}(20)$. First, find $x$ such that $g(x)=20$. Set $x^{3}-7 = 20$, then $x^{3}=27$, $x = 3$. So $f^{\prime}(20)=\frac{1}{g^{\prime}(3)}$. Substitute $x = 3$ into $g^{\prime}(x)$: $g^{\prime}(3)=3\times3^{2}=27$. So $f^{\prime}(20)=\frac{1}{27}$.

Step6: For problem 17

Find the derivative of $g(x)=\frac{5}{x + 3}=5(x + 3)^{-1}$. Using the chain - rule, $g^{\prime}(x)=-5(x + 3)^{-2}=-\frac{5}{(x + 3)^{2}}$. We want to find $f^{\prime}(\frac{1}{2})$. First, find $x$ such that $g(x)=\frac{1}{2}$. Set $\frac{5}{x + 3}=\frac{1}{2}$, then $x+3 = 10$, $x = 7$. So $f^{\prime}(\frac{1}{2})=\frac{1}{g^{\prime}(7)}$. Substitute $x = 7$ into $g^{\prime}(x)$: $g^{\prime}(7)=-\frac{5}{(7 + 3)^{2}}=-\frac{5}{100}=-\frac{1}{20}$. So $f^{\prime}(\frac{1}{2})=-20$.

Answer:

  1. $f^{\prime}(\frac{3\pi}{4})=\frac{1}{3\pi - 1}$
  2. $f^{\prime}(-10)=\frac{1}{23}$
  3. $f^{\prime}(4)=-4$
  4. $f^{\prime}(20)=\frac{1}{27}$
  5. $f^{\prime}(\frac{1}{2})=-20$