Question

f(x) is differentiable
x·f(x) + f(x) = \frac{d}{dx}(x^4 - x)
f(2) = 10
what is f(-1)?

Explanation:

Step1: Recognize the left - hand side as a product rule

We know that the product rule of differentiation states that \(\frac{d}{dx}(u(x)v(x))=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)\). Notice that the left - hand side of the equation \(x\cdot f^{\prime}(x)+f(x)\) can be written as \(\frac{d}{dx}(x\cdot f(x))\) because if we let \(u(x) = x\) and \(v(x)=f(x)\), then \(\frac{d}{dx}(x\cdot f(x))=x\cdot f^{\prime}(x)+1\cdot f(x)=x\cdot f^{\prime}(x)+f(x)\).

The right - hand side is \(\frac{d}{dx}(x^{4}-x)\). Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), we have \(\frac{d}{dx}(x^{4}-x)=4x^{3}-1\).

So our equation becomes \(\frac{d}{dx}(x\cdot f(x))=4x^{3}-1\).

Step2: Integrate both sides with respect to \(x\)

Integrating the left - hand side \(\int\frac{d}{dx}(x\cdot f(x))dx=x\cdot f(x)+C_1\) (by the fundamental theorem of calculus, the integral of a derivative is the original function).

Integrating the right - hand side \(\int(4x^{3}-1)dx\). Using the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we get \(\int(4x^{3}-1)dx = 4\cdot\frac{x^{4}}{4}-x+C=x^{4}-x + C\).

So we have \(x\cdot f(x)=x^{4}-x + C\).

Step3: Solve for \(f(x)\)

Divide both sides of the equation \(x\cdot f(x)=x^{4}-x + C\) by \(x\) (for \(x
eq0\)): \(f(x)=x^{3}-1+\frac{C}{x}\).

Step4: Find the value of \(C\) using \(f^{\prime}(2) = 10\)

First, find the derivative of \(f(x)\). If \(f(x)=x^{3}-1+\frac{C}{x}=x^{3}-1 + Cx^{-1}\), then \(f^{\prime}(x)=3x^{2}-Cx^{-2}=3x^{2}-\frac{C}{x^{2}}\).

We know that \(f^{\prime}(2) = 10\). Substitute \(x = 2\) into \(f^{\prime}(x)\):

\(f^{\prime}(2)=3(2)^{2}-\frac{C}{(2)^{2}}=12-\frac{C}{4}\)

Since \(f^{\prime}(2) = 10\), we set up the equation \(12-\frac{C}{4}=10\).

Subtract 12 from both sides: \(-\frac{C}{4}=10 - 12=-2\).

Multiply both sides by \(- 4\): \(C = 8\).

Step5: Find the function \(f(x)\)

Now that we know \(C = 8\), our function \(f(x)=x^{3}-1+\frac{8}{x}\) (for \(x
eq0\)).

Step6: Evaluate \(f(-1)\)

Substitute \(x=-1\) into \(f(x)\):

\(f(-1)=(-1)^{3}-1+\frac{8}{-1}=-1 - 1-8=-10\)

Answer:

\(f(-1)=-10\)