CALCU-Suppose f is a function which satisfies the

Suppose f is a function which satisfies the properties: f (x + y) = f (x) + f (y) + x2y + xy2 for all real numbers x and y, and f (x) lim = 1. x→0 x f ′(x) =

Solution by Steps

step 1

Given the function

f

satisfies

f(x + y) = f(x) + f(y) + x^2y + xy^2

for all real numbers

x

and

y

, and the limit

\lim_{x \to 0} \frac{f(x)}{x} = 1

. We need to find

\lim_{x \to 0} x f'(x)

step 2

Differentiate both sides of the given functional equation with respect to

x

:

f'(x + y) = f'(x) + 2xy + y^2

step 3

Set

y = 0

in the differentiated equation to get

f'(x) = f'(x) + 0 + 0

, which simplifies to

f'(x) = f'(x)

. This step confirms that the differentiation is consistent with the properties of

f

step 4

Now, take the limit as

x

approaches 0 of

x f'(x)

. Since

\lim_{x \to 0} \frac{f(x)}{x} = 1

, by the definition of the derivative,

f'(0) = 1

step 5

Therefore,

\lim_{x \to 0} x f'(x) = \lim_{x \to 0} x \cdot 1 = 0

Answer

\lim_{x \to 0} x f'(x) = 0

Key Concept

Limit of a Product and Derivative at a Point

Explanation

The limit of

x f'(x)

as

x

approaches 0 is found by recognizing that

f'(0)

is the derivative of

f(x)

at

x = 0

, which is given by the limit

\lim_{x \to 0} \frac{f(x)}{x}

. Since this limit equals 1,

f'(0)

must also equal 1, and thus the limit of the product

x f'(x)

is 0.

√ Let f (x) = 7 + 6 x. Then the expression f (x + h) − f (x) h can be written in the form A √ √ , Bx +Ch +( x) where A, B, and C are constants. (Note: It’s possible for one or more of these constants to be 0.) Find the constants. A= B= C= f (x + h) − f (x) Use your answer from above to find lim h→0 h. f (x + h) − f (x) = h→0 h lim Finally, find each of the following: f ′(1) = f ′(2) = f ′(3) =

Generated Graph

Solution by Steps

step 1

To find the constants A, B, and C, we simplify the expression

f(x + h) - f(x)

over

h

step 2

Using the asksia-ll calculator result, we have the simplified form:

6(h + x) - \frac{(6x + 7)}{h} + 7

step 3

As

h

approaches 0, the term

\frac{(6x + 7)}{h}

becomes undefined, so we only consider terms that do not involve

h

in the denominator

step 4

The terms without

h

in the denominator are

6x + 7

, which gives us A = 6 and B = 7. Since there is no

h

term left, C = 0

step 5

To find the limit as

h

approaches 0, we use the asksia-ll calculator result:

\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = 6

step 6

Finally, to find

f'(1)

,

f'(2)

, and

f'(3)

, we use the derivative of

f(x)

which is a constant 6

Answer

A = 6, B = 7, C = 0

\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = 6

f'(1) = 6, f'(2) = 6, f'(3) = 6

Key Concept

Simplification and differentiation of polynomial functions

Explanation

The expression

f(x + h) - f(x)

over

h

simplifies to a linear function in

x

and

h

, and the limit as

h

approaches 0 gives the derivative of

f(x)

. The derivative of a linear function is the coefficient of

x

.

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