Answer :
Below is a detailed explanation of how each part of the question is solved.
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1. Average Rate of Change of
[tex]$$f(x)=3x^4+2$$[/tex]
on the interval [tex]\([-2,3]\)[/tex]:
First, evaluate the function at the endpoints.
For [tex]\(x=3\)[/tex]:
[tex]\[
f(3)=3(3^4)+2=3(81)+2=243+2=245.
\][/tex]
For [tex]\(x=-2\)[/tex]:
[tex]\[
f(-2)=3((-2)^4)+2=3(16)+2=48+2=50.
\][/tex]
The average rate of change is given by
[tex]\[
\frac{f(3)-f(-2)}{3-(-2)}=\frac{245-50}{5}=\frac{195}{5}=39.
\][/tex]
──────────────────────────────
2. Instantaneous Rate of Change of
[tex]$$f(x)=2x^3-3$$[/tex]
at [tex]\(x=2\)[/tex]:
Differentiate the function:
[tex]\[
f'(x)=\frac{d}{dx}[2x^3-3]=6x^2.
\][/tex]
Now, evaluate at [tex]\(x=2\)[/tex]:
[tex]\[
f'(2)=6(2^2)=6(4)=24.
\][/tex]
──────────────────────────────
3. Derivative of
[tex]$$y=(x^3+3)^2$$[/tex]:
Apply the chain rule. Let
[tex]\[
u=x^3+3.
\][/tex]
Then,
[tex]\[
y=u^2 \quad \text{and} \quad \frac{dy}{dx}=2u\frac{du}{dx}.
\][/tex]
Differentiate [tex]\(u\)[/tex]:
[tex]\[
\frac{du}{dx}=3x^2.
\][/tex]
Thus,
[tex]\[
\frac{dy}{dx}=2(x^3+3)(3x^2)=6x^2(x^3+3).
\][/tex]
Distributing gives:
[tex]\[
6x^2\cdot x^3+6x^2\cdot 3=6x^5+18x^2.
\][/tex]
──────────────────────────────
4. Compute [tex]\(f'(2)\)[/tex] for
[tex]$$f(x)=x^3+3x$$[/tex]:
Differentiate:
[tex]\[
f'(x)=3x^2+3.
\][/tex]
Then, at [tex]\(x=2\)[/tex]:
[tex]\[
f'(2)=3(2^2)+3=3(4)+3=12+3=15.
\][/tex]
──────────────────────────────
5. Instantaneous Rate of Change of
[tex]$$f(x)=6x^2-3$$[/tex]
at [tex]\(x=2\)[/tex]:
Differentiate:
[tex]\[
f'(x)=12x.
\][/tex]
Then,
[tex]\[
f'(2)=12(2)=24.
\][/tex]
──────────────────────────────
6. Find [tex]\(F'(1)\)[/tex] for
[tex]$$F(x)=f(x^2+4x) \cdot g(3x+1)$$[/tex]
given:
[tex]\[
f(5)=-2,\quad f'(5)=2,\quad g(4)=3,\quad g'(4)=5.
\][/tex]
Define auxiliary functions:
[tex]\[
u(x)=x^2+4x,\quad v(x)=3x+1.
\][/tex]
Differentiate these:
[tex]\[
u'(x)=2x+4\quad\text{and}\quad v'(x)=3.
\][/tex]
At [tex]\(x=1\)[/tex]:
[tex]\[
u(1)=1^2+4(1)=1+4=5,\quad u'(1)=2(1)+4=6,
\][/tex]
[tex]\[
v(1)=3(1)+1=4,\quad v'(1)=3.
\][/tex]
Apply the product rule combined with the chain rule:
[tex]\[
F'(1)=f'(u(1))\cdot u'(1)\cdot g(v(1)) + f(u(1))\cdot g'(v(1))\cdot v'(1).
\][/tex]
Substitute the known values:
[tex]\[
F'(1)=2\cdot6\cdot3 + (-2)\cdot5\cdot3=36-30=6.
\][/tex]
──────────────────────────────
7. Find [tex]\(f'(1)\)[/tex] for
[tex]$$f(x)=\frac{g(x)}{x+1}+(g(x))^2$$[/tex]
with
[tex]\[
g(1)=8,\quad g'(1)=2.
\][/tex]
Differentiate the first term using the quotient rule:
[tex]\[
\frac{d}{dx}\left[\frac{g(x)}{x+1}\right]=\frac{g'(x)(x+1)-g(x)}{(x+1)^2}.
\][/tex]
Differentiate the second term using the chain rule:
[tex]\[
\frac{d}{dx}\left[(g(x))^2\right]=2g(x)g'(x).
\][/tex]
Thus,
[tex]\[
f'(x)=\frac{g'(x)(x+1)-g(x)}{(x+1)^2}+2g(x)g'(x).
\][/tex]
Evaluate at [tex]\(x=1\)[/tex]:
[tex]\[
\frac{g'(1)(1+1)-g(1)}{(1+1)^2}+\;2g(1)g'(1)=\frac{2\cdot2-8}{4}+2\cdot8\cdot2.
\][/tex]
Compute the first term:
[tex]\[
\frac{4-8}{4}=\frac{-4}{4}=-1.
\][/tex]
Compute the second term:
[tex]\[
2\cdot8\cdot2=32.
\][/tex]
Then,
[tex]\[
f'(1)=-1+32=31.
\][/tex]
──────────────────────────────
8. Identify the False Statement about Derivatives:
Among the given options, consider the following statements:
- The derivative of a function at a point is the instantaneous rate of change of the function at that point.
- The derivative of a function at a point is the slope of the tangent line to its graph.
- The derivative is defined as the limit of the ratio of the difference of function values to the difference in the input values.
The false statement is the one that claims the derivative at a point is equal to the area of the region enclosed by the graph of the function. This is not true.
──────────────────────────────
Final Answers:
1. Average rate of change: [tex]\(\boxed{39}\)[/tex].
2. Instantaneous rate of change for [tex]\(f(x)=2x^3-3\)[/tex] at [tex]\(x=2\)[/tex]: [tex]\(\boxed{24}\)[/tex].
3. Derivative of [tex]\(y=(x^3+3)^2\)[/tex]: [tex]\(\boxed{6x^5+18x^2}\)[/tex].
4. [tex]\(f'(2)\)[/tex] for [tex]\(f(x)=x^3+3x\)[/tex]: [tex]\(\boxed{15}\)[/tex].
5. Instantaneous rate of change for [tex]\(f(x)=6x^2-3\)[/tex] at [tex]\(x=2\)[/tex]: [tex]\(\boxed{24}\)[/tex].
6. [tex]\(F'(1)\)[/tex] for [tex]\(F(x)=f(x^2+4x) \cdot g(3x+1)\)[/tex]: [tex]\(\boxed{6}\)[/tex].
7. [tex]\(f'(1)\)[/tex] for [tex]\(f(x)=\frac{g(x)}{x+1}+(g(x))^2\)[/tex]: [tex]\(\boxed{31}\)[/tex].
8. The statement that is NOT true about derivatives is the one stating that the derivative of a function at a point is equal to the area of the region enclosed by the graph; that corresponds to option [tex]\(\boxed{B}\)[/tex].
──────────────────────────────
1. Average Rate of Change of
[tex]$$f(x)=3x^4+2$$[/tex]
on the interval [tex]\([-2,3]\)[/tex]:
First, evaluate the function at the endpoints.
For [tex]\(x=3\)[/tex]:
[tex]\[
f(3)=3(3^4)+2=3(81)+2=243+2=245.
\][/tex]
For [tex]\(x=-2\)[/tex]:
[tex]\[
f(-2)=3((-2)^4)+2=3(16)+2=48+2=50.
\][/tex]
The average rate of change is given by
[tex]\[
\frac{f(3)-f(-2)}{3-(-2)}=\frac{245-50}{5}=\frac{195}{5}=39.
\][/tex]
──────────────────────────────
2. Instantaneous Rate of Change of
[tex]$$f(x)=2x^3-3$$[/tex]
at [tex]\(x=2\)[/tex]:
Differentiate the function:
[tex]\[
f'(x)=\frac{d}{dx}[2x^3-3]=6x^2.
\][/tex]
Now, evaluate at [tex]\(x=2\)[/tex]:
[tex]\[
f'(2)=6(2^2)=6(4)=24.
\][/tex]
──────────────────────────────
3. Derivative of
[tex]$$y=(x^3+3)^2$$[/tex]:
Apply the chain rule. Let
[tex]\[
u=x^3+3.
\][/tex]
Then,
[tex]\[
y=u^2 \quad \text{and} \quad \frac{dy}{dx}=2u\frac{du}{dx}.
\][/tex]
Differentiate [tex]\(u\)[/tex]:
[tex]\[
\frac{du}{dx}=3x^2.
\][/tex]
Thus,
[tex]\[
\frac{dy}{dx}=2(x^3+3)(3x^2)=6x^2(x^3+3).
\][/tex]
Distributing gives:
[tex]\[
6x^2\cdot x^3+6x^2\cdot 3=6x^5+18x^2.
\][/tex]
──────────────────────────────
4. Compute [tex]\(f'(2)\)[/tex] for
[tex]$$f(x)=x^3+3x$$[/tex]:
Differentiate:
[tex]\[
f'(x)=3x^2+3.
\][/tex]
Then, at [tex]\(x=2\)[/tex]:
[tex]\[
f'(2)=3(2^2)+3=3(4)+3=12+3=15.
\][/tex]
──────────────────────────────
5. Instantaneous Rate of Change of
[tex]$$f(x)=6x^2-3$$[/tex]
at [tex]\(x=2\)[/tex]:
Differentiate:
[tex]\[
f'(x)=12x.
\][/tex]
Then,
[tex]\[
f'(2)=12(2)=24.
\][/tex]
──────────────────────────────
6. Find [tex]\(F'(1)\)[/tex] for
[tex]$$F(x)=f(x^2+4x) \cdot g(3x+1)$$[/tex]
given:
[tex]\[
f(5)=-2,\quad f'(5)=2,\quad g(4)=3,\quad g'(4)=5.
\][/tex]
Define auxiliary functions:
[tex]\[
u(x)=x^2+4x,\quad v(x)=3x+1.
\][/tex]
Differentiate these:
[tex]\[
u'(x)=2x+4\quad\text{and}\quad v'(x)=3.
\][/tex]
At [tex]\(x=1\)[/tex]:
[tex]\[
u(1)=1^2+4(1)=1+4=5,\quad u'(1)=2(1)+4=6,
\][/tex]
[tex]\[
v(1)=3(1)+1=4,\quad v'(1)=3.
\][/tex]
Apply the product rule combined with the chain rule:
[tex]\[
F'(1)=f'(u(1))\cdot u'(1)\cdot g(v(1)) + f(u(1))\cdot g'(v(1))\cdot v'(1).
\][/tex]
Substitute the known values:
[tex]\[
F'(1)=2\cdot6\cdot3 + (-2)\cdot5\cdot3=36-30=6.
\][/tex]
──────────────────────────────
7. Find [tex]\(f'(1)\)[/tex] for
[tex]$$f(x)=\frac{g(x)}{x+1}+(g(x))^2$$[/tex]
with
[tex]\[
g(1)=8,\quad g'(1)=2.
\][/tex]
Differentiate the first term using the quotient rule:
[tex]\[
\frac{d}{dx}\left[\frac{g(x)}{x+1}\right]=\frac{g'(x)(x+1)-g(x)}{(x+1)^2}.
\][/tex]
Differentiate the second term using the chain rule:
[tex]\[
\frac{d}{dx}\left[(g(x))^2\right]=2g(x)g'(x).
\][/tex]
Thus,
[tex]\[
f'(x)=\frac{g'(x)(x+1)-g(x)}{(x+1)^2}+2g(x)g'(x).
\][/tex]
Evaluate at [tex]\(x=1\)[/tex]:
[tex]\[
\frac{g'(1)(1+1)-g(1)}{(1+1)^2}+\;2g(1)g'(1)=\frac{2\cdot2-8}{4}+2\cdot8\cdot2.
\][/tex]
Compute the first term:
[tex]\[
\frac{4-8}{4}=\frac{-4}{4}=-1.
\][/tex]
Compute the second term:
[tex]\[
2\cdot8\cdot2=32.
\][/tex]
Then,
[tex]\[
f'(1)=-1+32=31.
\][/tex]
──────────────────────────────
8. Identify the False Statement about Derivatives:
Among the given options, consider the following statements:
- The derivative of a function at a point is the instantaneous rate of change of the function at that point.
- The derivative of a function at a point is the slope of the tangent line to its graph.
- The derivative is defined as the limit of the ratio of the difference of function values to the difference in the input values.
The false statement is the one that claims the derivative at a point is equal to the area of the region enclosed by the graph of the function. This is not true.
──────────────────────────────
Final Answers:
1. Average rate of change: [tex]\(\boxed{39}\)[/tex].
2. Instantaneous rate of change for [tex]\(f(x)=2x^3-3\)[/tex] at [tex]\(x=2\)[/tex]: [tex]\(\boxed{24}\)[/tex].
3. Derivative of [tex]\(y=(x^3+3)^2\)[/tex]: [tex]\(\boxed{6x^5+18x^2}\)[/tex].
4. [tex]\(f'(2)\)[/tex] for [tex]\(f(x)=x^3+3x\)[/tex]: [tex]\(\boxed{15}\)[/tex].
5. Instantaneous rate of change for [tex]\(f(x)=6x^2-3\)[/tex] at [tex]\(x=2\)[/tex]: [tex]\(\boxed{24}\)[/tex].
6. [tex]\(F'(1)\)[/tex] for [tex]\(F(x)=f(x^2+4x) \cdot g(3x+1)\)[/tex]: [tex]\(\boxed{6}\)[/tex].
7. [tex]\(f'(1)\)[/tex] for [tex]\(f(x)=\frac{g(x)}{x+1}+(g(x))^2\)[/tex]: [tex]\(\boxed{31}\)[/tex].
8. The statement that is NOT true about derivatives is the one stating that the derivative of a function at a point is equal to the area of the region enclosed by the graph; that corresponds to option [tex]\(\boxed{B}\)[/tex].
