Question

let $f$ be a differentiable function such that $f(1)=2$ and $f(x)=sqrt{x^{2}+2cos x + 3}$. what is the value of $f(4)$?
(a) - 6.790
(b) 8.790
(c) 10.790
(d) 12.996

Explanation:

Step1: Use the fundamental theorem of calculus

By the fundamental theorem of calculus, $f(4)-f(1)=\int_{1}^{4}f^{\prime}(x)dx$. We know that $f(1) = 2$ and $f^{\prime}(x)=\sqrt{x^{2}+2\cos x + 3}$, so $f(4)=2+\int_{1}^{4}\sqrt{x^{2}+2\cos x + 3}dx$.

Step2: Approximate the integral

We can use a numerical - integration method (such as the trapezoidal rule or a calculator with integral - calculation capabilities). Using a calculator with integral - calculation functions, $\int_{1}^{4}\sqrt{x^{2}+2\cos x + 3}dx\approx8.790$.

Step3: Calculate $f(4)$

$f(4)=2+\int_{1}^{4}\sqrt{x^{2}+2\cos x + 3}dx\approx2 + 8.790=10.790$.

Answer:

C. 10.790