- #1

karush

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The function f is defined by

$$f(x)=\sqrt{25-x^2},\quad -5\le x \le 5$$

(a) Find $f'(x)$ apply chain rule

$$

\dfrac{d}{dx}(25-x^2)^{1/2}

=\dfrac{1}{2}(25-x^2)^{-1/2}2x

=-\frac{x}{\sqrt{25-x^2}}$$

(b) Write an equation for the tangent line to the graph of f at $x=-3$

$$f'(-3)=-\frac{3}{\sqrt{25-(3)^2}}

=-\dfrac{3}{4}=m$$

then f(-3)=4 $y=mx+b$ so $y=-\dfrac{3}{4}(x+3)+4$

(c) Let g be the function defined by $\biggr\{\begin{array}{ll}

f(x) &\textit{for } -5\le x \ge 5\\

x+7 &\textit{for } 3\le x\le 5

\end{array}$

Is g continuous at $x=-3$ (d) Find the value of

$\displaystyle

\int_0^5 x \sqrt{25-x^2}\, dx$

---------------------------------------------------------------

ok i think a and b are okbut (c) x+7 is not a tangent line but looks continuous by the inequalities(d) I assume they tossed in the x for a u substitution method.

View attachment 9331

$$f(x)=\sqrt{25-x^2},\quad -5\le x \le 5$$

(a) Find $f'(x)$ apply chain rule

$$

\dfrac{d}{dx}(25-x^2)^{1/2}

=\dfrac{1}{2}(25-x^2)^{-1/2}2x

=-\frac{x}{\sqrt{25-x^2}}$$

(b) Write an equation for the tangent line to the graph of f at $x=-3$

$$f'(-3)=-\frac{3}{\sqrt{25-(3)^2}}

=-\dfrac{3}{4}=m$$

then f(-3)=4 $y=mx+b$ so $y=-\dfrac{3}{4}(x+3)+4$

(c) Let g be the function defined by $\biggr\{\begin{array}{ll}

f(x) &\textit{for } -5\le x \ge 5\\

x+7 &\textit{for } 3\le x\le 5

\end{array}$

Is g continuous at $x=-3$ (d) Find the value of

$\displaystyle

\int_0^5 x \sqrt{25-x^2}\, dx$

---------------------------------------------------------------

ok i think a and b are okbut (c) x+7 is not a tangent line but looks continuous by the inequalities(d) I assume they tossed in the x for a u substitution method.

View attachment 9331

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