Question

1. if (f(x)=int_{1}^{x}tcos(t)dt) and (g(x)=x^{2}), then what is ((fcirc g)(x)?

(a) ((fcirc g)(x)=int_{1}^{x^{2}}x^{2}cos(x^{2})dx
(b) ((fcirc g)(x)=int_{1}^{x^{2}}tcos(t^{2})dt
(c) ((fcirc g)(x)=int_{1}^{x^{2}}x^{2}cos(x^{2})dx
(d) ((fcirc g)(x)=int_{1}^{x^{2}}tcos(t)dt

1. the tangent line to (f(x)=(1 + x)^{\frac{1}{4}}) at (x = 0) is (y=1+\frac{1}{4}x). using this tangent line to approximate ((1.1)^{\frac{1}{4}}) gives:

(a) ((1.1)^{\frac{1}{4}}approx\frac{37}{36}
(b) ((1.1)^{\frac{1}{4}}approx\frac{39}{40}
(c) ((1.1)^{\frac{1}{4}}approx1
(d) ((1.1)^{\frac{1}{4}}approx\frac{41}{40}
(e) ((1.1)^{\frac{1}{4}}approx\frac{5}{4})

Explanation:

Step1: Recall the definition of composition of functions

By the definition of $(f\circ g)(x)=f(g(x))$. Given $f(x)=\int_{1}^{x}t\cos(t)dt$ and $g(x) = x^{2}$, we substitute $g(x)$ into $f(x)$.

Step2: Make the substitution

We get $(f\circ g)(x)=\int_{1}^{g(x)}t\cos(t)dt=\int_{1}^{x^{2}}t\cos(t)dt$.

Answer:

(b) $(f\circ g)(x)=\int_{1}^{x^{2}}t\cos(t)dt$