Question

Find the number \(\mathrm{K}, \mathrm{L}\) and \(\mathrm{M}\) such that the function of the form \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{Kx}^{2}+\mathrm{L}}{\mathrm{x}-1}+\mathrm{Mx}\) satisfies the conditions \(f(2)=23, f^{\prime}(0)=4\) and \(\int_{-1}^{0}(x-1) f(x) d x=\frac{37}{6}\)

Short answer

Answer: The values for K, L, and M are K = 1, L = \(\frac{11}{3}\), and M = \(\frac{23}{3}\).

Step-by-step solution

Find the derivative of f(x)

First, we need to find the derivative \(f'(x)\) of the given function \(f(x)=\frac{Kx^2+L}{x-1}+Mx\). Using the quotient rule, the derivative of the first part is: \( \frac{d}{dx}(\frac{Kx^2+L}{x-1})=\frac{(2Kx)(x-1)-(Kx^2+L)}{(x-1)^2}\). Now, differentiate the second part: \( \frac{d}{dx}Mx=M \) So, the derivative of f(x) is: \(f'(x)=\frac{(2Kx)(x-1)-(Kx^2+L)}{(x-1)^2}+M\)

Use conditions 1 and 2 to find K and L

Now, we'll use the conditions given to find the values of K and L. 1. \(f(2)=23 \Rightarrow \frac{K(2)^2+L}{2-1}+2M=23 \Rightarrow 4K+L+2M=23\) 2. \(f'(0)=4 \Rightarrow \frac{(2K(0))(0-1)-(K(0)^2+L)}{(0-1)^2}+M=4 \Rightarrow -L+M=4\) We have the system of equations: a) \(4K+L+2M=23\) b) \(-L+M=4\) We'll need one more equation in K, L, and M to solve this system.

Use condition 3 to find the value of M

The third condition is the integral: \(\int_{-1}^0 (x-1)f(x)dx = \frac{37}{6}\) Let's evaluate this integral and express it in terms of K, L, and M: \(\int_{-1}^0 (x-1)\left(\frac{Kx^2+L}{x-1}+Mx\right)dx = \int_{-1}^0 (Kx^2+L+Mx^2(x-1))dx = \frac{37}{6}\) Now, integrate the function with respect to x: \(\int (Kx^3+Lx+Mx^3(x-1))dx = \frac{1}{4}Kx^4+\frac{1}{2}Lx^2+\frac{1}{4}Mx^4-\frac{1}{4}Mx^3 \Big |_{-1}^0 = \frac{37}{6}\) Evaluate the integral at the limits: \([(\frac{1}{4}K(0)^4+\frac{1}{2}L(0)^2+\frac{1}{4}M(0)^4-\frac{1}{4}M(0)^3) - (\frac{1}{4}K(-1)^4+\frac{1}{2}L(-1)^2+\frac{1}{4}M(-1)^4-\frac{1}{4}M(-1)^3)] = \frac{37}{6} \Rightarrow \frac{1}{4}K+\frac{1}{2}L+\frac{1}{4}M-\frac{1}{4}M=\frac{37}{6}\) So, we get the third equation from the integral condition: c) \(\frac{1}{4}K+\frac{1}{2}L= \frac{37}{6}\) Now, we have a system of three equations in terms of K, L, and M: a) \(4K+L+2M=23\) b) \(-L+M=4\) c) \(\frac{1}{4}K+\frac{1}{2}L= \frac{37}{6}\)

Solve the system of equations for K, L, and M

Now, we'll solve this system of equations to find the values of K, L, and M: a) \(4K+L+2M=23\) b) \(-L+M=4 \Rightarrow M=L+4\) c) \(\frac{1}{4}K+\frac{1}{2}L= \frac{37}{6}\) Substituting b) into a): \(4K+L+2(L+4)=23 \Rightarrow 4K+3L=15\) Solve for L: \(L=\frac{15-4K}{3}\) Substituting this into c) and solving for K: \(\frac{1}{4}K+\frac{1}{2}\left(\frac{15-4K}{3}\right)=\frac{37}{6} \Rightarrow K=1\) Now, we can find the value of L: \(L=\frac{15-4(1)}{3}=\frac{11}{3}\) And finally, find the value of M: \(M=L+4=\frac{11}{3}+4=\frac{23}{3}\) So, the values for K, L, and M are K=1, L=\(\frac{11}{3}\), and M=\(\frac{23}{3}\).