CALCU-7. Discuss the continuity of the function\[f

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Solution by Steps

step 1

To discuss the continuity of the function

f(x) = \begin{cases} x^2 - 3x &amp; \text{if } x &lt; 2 \\ 4 + 2x &amp; \text{if } x \geq 2 \end{cases}

, we first need to check the continuity on the open interval 0 < x < 2

step 2

The function

f(x) = x^2 - 3x

is a polynomial for x < 2 , which is continuous everywhere in its domain. Thus, it is continuous on

(0, 2)

step 3

Next, we check the continuity at the point

x = 2

for the closed interval

[0, 2]

. We need to evaluate

f(2)

using the second piece of the function:

f(2) = 4 + 2(2) = 8

step 4

We also need to find the limit of

f(x)

as

x

approaches 2 from the left:

\lim_{x \to 2^-} f(x) = 2^2 - 3(2) = 4 - 6 = -2

. Since

\lim_{x \to 2^-} f(x) \neq f(2)

, the function is not continuous at

x = 2

step 5

Therefore,

f(x)

is continuous on the open interval

(0, 2)

but not continuous on the closed interval

[0, 2]

due to the discontinuity at

x = 2

Answer

The function is continuous on

(0, 2)

but not on

[0, 2]

due to a discontinuity at

x = 2

.

Key Concept

Continuity of piecewise functions

Explanation

A function is continuous at a point if the limit as you approach that point equals the function's value at that point. In this case, the limit from the left does not equal the function value at

x = 2

.

Solution by Steps

step 1

To find the derivative of

f(x) = \frac{4}{\sqrt{1-x}}

using the definition of derivative, we use the limit definition:

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

step 2

Substituting

a

into the limit, we have

f'(a) = \lim_{h \to 0} \frac{\frac{4}{\sqrt{1-(a+h)}} - \frac{4}{\sqrt{1-a}}}{h}

step 3

Simplifying the expression, we find a common denominator and evaluate the limit as

h

approaches 0

step 4

The derivative

f'(a)

is calculated to be

\frac{2}{(1-a)^{3/2}}

step 5

For

f(t) = \frac{2t+1}{t+3}

, we apply the same limit definition:

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

step 6

Substituting

a

into the limit gives

f'(a) = \lim_{h \to 0} \frac{\frac{2(a+h)+1}{(a+h)+3} - \frac{2a+1}{a+3}}{h}

step 7

After simplifying, we find

f'(a) = \frac{5}{(a+3)^2}

step 8

For

f(x) = \sqrt{1-2x}

, we again use the limit definition:

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

step 9

Substituting

a

gives

f'(a) = \lim_{h \to 0} \frac{\sqrt{1-2(a+h)} - \sqrt{1-2a}}{h}

step 10

Simplifying this expression leads to

f'(a) = \frac{-1}{\sqrt{1-2a}}

step 11

To find the equation of the tangent line to

y = 3x^2 - x^3

at the point

(1,2)

, we first find

f'(1)

using the definition of derivative

step 12

We calculate

f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{(3(1+h)^2 - (1+h)^3) - 2}{h}

step 13

After simplifying, we find

f'(1) = 3

step 14

The equation of the tangent line at the point

(1,2)

is given by

y - 2 = 3(x - 1)

, which simplifies to

y = 3x - 1

Answer

The derivatives are

f'(a) = \frac{2}{(1-a)^{3/2}}

,

f'(a) = \frac{5}{(a+3)^2}

,

f'(a) = \frac{-1}{\sqrt{1-2a}}

, and the tangent line is

y = 3x - 1

Key Concept

The definition of the derivative provides a way to find the slope of a function at a given point.

Explanation

By applying the limit definition of the derivative, we can find the slopes of the given functions and the equation of the tangent line at a specific point.

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