*Question:*

Let f be a function defined by \(\left\{\begin{matrix}

1-2 sin x & for x \leq 0 & & \\e^{-4x}

&for x > 0. & &

\end{matrix}\right.\)

**(a) Show that f is continuous at x = 0.**

(b) For x ≠ 0, express f'(x) as a piecewise-defined function. Find the value of x for which f'(x) = -3.

(c) Find the average value of f on the interval [-1, 1] .

**▶️Answer/Explanation**

Ans:

(a)

To be continuous

i) f(0) = 1

ii) \(\lim_{x\rightarrow 0} 1 -2 sin x = \lim_{x\rightarrow 0}e^{-4x}\)

1 = 1 ∴ \(\lim_{x\rightarrow 0^{-}} f = \lim_{x\rightarrow 0^{+}}f\)

∴ \(\lim_{x\rightarrow 0} f = 1\)

iii) f(0) = \(\lim_{x\rightarrow 0} f = 1\)

∴ f is continuous for all values of x.

(b)

\(f'(x)=\left\{\begin{matrix}

-2 cos x &, &x<0 & \\-4e^{-4x}

&, &x>0 &

\end{matrix}\right.\)

\({f}'(x) = 3\)

Since -2 cos x oscillates between -2 and 2 there will be no such value in this function such that f'(x) = -3

but \({f}'(x) =- 3\)

-4e ^{-4x} = -3

e ^{-4x} = 3/4

-4x = In (3/4) \(\therefore = \frac{-1}{4}In\left ( \frac{3}{4} \right )\therefore f’\left ( \frac{-1}{4}In\left ( \frac{3}{4} \right ) \right )= -3\)

(c)

\(f_{avg} = \frac{1}{1+1}\int_{-1}^{1}f(x)dx\)

\(= \frac{1}{2}\left [ \int_{-1}^{0} 1-2 sinx dx + \int_{0}^{1}e^{-4x}dx\right ]\)

\(= \frac{1}{2}\left [ [x+2cosx]_{-1}^{0}+[\frac{-1}{4}e^{-4x}]_{0}^{1} \right ]\)

\(= \frac{1}{2}\left [ 2+1-2cos(-1)\frac{-e^{-4}}{4} +\frac{1}{4}\right ]\)

\(=\frac{-\left [ 3-2cos(-1) \right ]+\left [ \frac{-1}{4e^{4}} +\frac{1}{4}\right ]}{2}\)

*Question:*

The function f is defined by \(f(x)=\sqrt{25-x^{2}}\) for -5 ≤ x ≤ 5.

(a) Find f'(x).

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

**(c) Let g be the function defined by \(g(x)=\left\{\begin{matrix}f(x) &for -5\leq x\leq -3 & & \\x+7 &for -3<x\leq 5. & & \end{matrix}\right.\)**

Is g continuous at x = -3 ? Use the definition of continuity to explain your answer

(d) Find the value of \(\int_{0}^{5}x\sqrt{25-x^{2}}dx.\)

**▶️Answer/Explanation**

Ans:

(a)

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

\(f'(x)=\frac{1.-2x}{2\sqrt{25-x^{2}}}=\frac{-x}{\sqrt{25-x^{2}}}\)

(b)

\(f'(-3)=\frac{3}{\sqrt{25-9}}=\frac{3}{\sqrt{16}}=\frac{3}{4}\) \(f(-3)=\sqrt{25-9}= \sqrt{16}=4\)

\(y = \frac{3}{4}(x+3)+4\)

(c)

\(\lim_{x\rightarrow -3^{-}}g(x)=f(-3)=4\) therefore \(\lim_{x\rightarrow -3}g(x)=4\)

\(\lim_{x\rightarrow -3^{+}}g(x)=-3+7=4\)

g(-3) = f(-3) = 4 \(\lim_{x\rightarrow -3}g(x)\) exists and is equal to g(-3)

g is continuous at x = -3

(d)

u = 25-x^{2} \(\frac{-1}{2}\int_{25}^{0}4^{1/2}du = \frac{1}{2}\int_{0}^{25}4^{1/2}du = \frac{1}{2}.\frac{24}{3}^{3/2}\int_{0}^{25}\)

\(\frac{-1}{2}du = +2xdx\)

\(= \frac{1}{3}.25^{3/2}=\frac{125}{3}\)

*Question:*

The function f is differentiable on the closed interval [−6, 5 ] and satisfies f (−2) = 7. The graph of f’, the derivative of f, consists of a semicircle and three line segments, as shown in the figure above.

(a) Find the values of f (−6) and f (5).

(b) On what intervals is f increasing? Justify your answer.

(c) Find the absolute minimum value of f on the closed interval [−6, 5 ]. Justify your answer.**(d) For each of f”(-5) and f “(3 ), find the value or explain why it does not exist.**

**▶️Answer/Explanation**

Ans:

(a)

\(f(-6)=\left ( \int_{-2}^{-6}f'(x)dx \right )+f(-2)\)

f(-6) = 3

\(f(5)=f(-2)+\int_{-2}^{5}f'(x)dx\)

f(5) = 10 – 2π

(b)

f is increasing on x = [-6, -2]

u [2, 5], since f’> 0 on the interval x ∈ [-6, -2] v[2, 5]

(c)

The absolute minimum endpoints of f or [-6, 5] is f(-6) = 3

7 – 2 π , since f(5) = 10 – 2π

f(2) < f (5) and f(-6) critical points

the endpoints f’ = 0

and f(2) < f(-2) the f(-2) = 7

other critical points, f(2) = 7.2 π

by EVT

(d)

\(f”(-5)=\frac{-1}{2}\)

f”(3) = DNE, as the

\(\lim_{x\rightarrow 3^{+}}\frac{f'(x)-2}{x-3}\neq \lim_{x\rightarrow 3^{-}}\frac{f'(x)-2}{x-3}\)

Therefore it is impossible to take a derivative at x = 3 in f’

*Question:*

The function f is differentiable on the closed interval [−6, 5 ] and satisfies f (−2) = 7. The graph of f’, the derivative of f, consists of a semicircle and three line segments, as shown in the figure above.

(a) Find the values of f (−6) and f (5).

(b) On what intervals is f increasing? Justify your answer.

(c) Find the absolute minimum value of f on the closed interval [−6, 5 ]. Justify your answer.**(d) For each of f”(-5) and f “(3 ), find the value or explain why it does not exist.**

**▶️Answer/Explanation**

Ans:

(a)

\(f(-6)=\left ( \int_{-2}^{-6}f'(x)dx \right )+f(-2)\)

f(-6) = 3

\(f(5)=f(-2)+\int_{-2}^{5}f'(x)dx\)

f(5) = 10 – 2π

(b)

f is increasing on x = [-6, -2]

u [2, 5], since f’> 0 on the interval x ∈ [-6, -2] v[2, 5]

(c)

The absolute minimum endpoints of f or [-6, 5] is f(-6) = 3

7 – 2 π , since f(5) = 10 – 2π

f(2) < f (5) and f(-6) critical points

the endpoints f’ = 0

and f(2) < f(-2) the f(-2) = 7

other critical points, f(2) = 7.2 π

by EVT

(d)

\(f”(-5)=\frac{-1}{2}\)

f”(3) = DNE, as the

\(\lim_{x\rightarrow 3^{+}}\frac{f'(x)-2}{x-3}\neq \lim_{x\rightarrow 3^{-}}\frac{f'(x)-2}{x-3}\)

Therefore it is impossible to take a derivative at x = 3 in f’

*Question:*

Functions f, g, and h are twice-differentiable functions with g(2) = h(2) = 4. The line \(y = 4+\frac{2}{3}(x-2)\) is tangent to both the graph of g at x = 2 and the graph of h at x = 2.

(a) Find h'(2).

(b) Let a be the function given by a(x) = 3x^{3}h(x). Write an expression for a'(x). Find a'(2).

(c) The function h satisfies \(h(x)=\frac{x^{2}-4}{1-(f(x))^{3}}\) for x ≠ 2. It is known that \(\lim_{x\rightarrow 2}h(x)\) can be evaluated using L’Hospital’s Rule. Use \(\lim_{x\rightarrow 2}h(x)\) to find f(2) and f'(2 ). Show the work that leads to your answers.**(d) It is known that g(x) ≤ h(x) for 1 < x < 3. Let k be a function satisfying g(x) ≤ k(x) ≤ h(x) for 1 < x < 3. Is k continuous at x = 2 ? Justify your answer.**

**▶️Answer/Explanation**

Ans:

(a)

\(y = \frac{2}{3}(x-2)+4\)

point (2, 4) slope 2/3

h'(2) = 2/3

(b)

a'(x) = h'(x).3x^{3} + 9x^{2} h(x)

a'(2) = h'(2) . 3(2)^{3} + 9(2)^{2} h(2)

\(a'(2)=\frac{2}{3}\cdot 3\cdot 2^{3}+9\cdot 2^{2}\cdot 4\)

(c)

\(\lim_{x\rightarrow 2}\frac{x^{2}-4}{1-(f(x))^{3}}= \lim_{x\rightarrow 2}\frac{2x}{-3(f(x))^{2}.f'(x)}\) h (2) = 4

\(\therefore \lim_{x\rightarrow 2} h(x)\) must equal 4

\(\therefore \lim_{x\rightarrow 2} x^{2}-4=0\)

\(\lim_{x\rightarrow 2} 1-(f(x))^{3}=0\) if f(2) = 1 \(=\frac{2(2)}{-3(1)^{2}\cdot f'(2)}\)

∴ Indeterminate form 0/0 \(=\frac{4}{-3\cdot f'(2)}\) f'(2) must equal -1/3 so that \(\lim_{x\rightarrow 2} h(x) = 4\)

and L’Hopital’s rule applies \(=\frac{4}{-3\cdot -1/3}\)

Such that f(2) = 1 = 4/1

\(\lim_{x\rightarrow 2} h(x) = 4\) such that f(2) = 1 and f'(2) = -1/3

(d)

2 and h given as twice – differentiable ∴ g and h are continuous for 1 < x < 3

g(x) ≤ k(x) ≤ h(x)

\(\lim_{x\rightarrow 2} g(x) = 4\) because g continuous and g(2) = 4

\(\lim_{x\rightarrow 2} h(x) = 4\) because h continuous and h(2) = 4

By squeeze theorem, \(\lim_{x\rightarrow 2} k(x) = 4\)

k is between or equal to g and h for 1 < x < 3, given g(2) = 4 h (2) , meaning k(2) must equal 4, and \(\lim_{x\rightarrow 2} k(x) = 4\) so k(x) is continuous at x = 2