question 6 (1 point)\nthe linear approximation to f(x) = e^x sin x+cos x at the point x = 0 is given by…

Explanation:

Step1: Recall linear - approximation formula

The linear approximation of a function $y = f(x)$ at $x = a$ is given by $L(x)=f(a)+f^{\prime}(a)(x - a)$. Here $a = 0$.

Step2: Find $f(0)$

Given $f(x)=e^{x}\sin x+\cos x$, then $f(0)=e^{0}\sin(0)+\cos(0)=0 + 1=1$.

Step3: Find the derivative of $f(x)$

Using the product - rule $(uv)^\prime=u^\prime v+uv^\prime$ where $u = e^{x}$ and $v=\sin x$. The derivative of $e^{x}\sin x$ is $e^{x}\sin x+e^{x}\cos x$, and the derivative of $\cos x$ is $-\sin x$. So $f^{\prime}(x)=e^{x}\sin x+e^{x}\cos x-\sin x$.

Step4: Find $f^{\prime}(0)$

Substitute $x = 0$ into $f^{\prime}(x)$: $f^{\prime}(0)=e^{0}\sin(0)+e^{0}\cos(0)-\sin(0)=0 + 1-0 = 1$.

Step5: Find the linear approximation $L(x)$

Since $a = 0$, $f(0)=1$ and $f^{\prime}(0)=1$, then $L(x)=f(0)+f^{\prime}(0)(x - 0)=1+1\times x=x + 1$.

Answer:

$L(x)=x + 1$