Aleena Kaiser

2022-04-30

Antiderivatives, working backwards.

Sorry if my title is a bit confusing. I have the question:

A function f defined for all real numbers satisfies:

${f}^{\prime}(1)=1$, $f(0)=4$, and ${f}^{\u2033}(x)=12{x}^{2}-12x$. Find the value of f(2).

I left out some unimportant parts which I should be capable of doing myself after I understand this bit, my question is am I doing this correctly?

${f}^{\prime}(x)=4{x}^{3}-6{x}^{2}+C$

$4(1{)}^{3}-6(1{)}^{2}+C=1$

$4-6+C=1$

$C=3$

$f(x)={x}^{4}-2{x}^{3}+C$

$f(0)={0}^{4}-2(0{)}^{3}+C=4$

$C=4$

$f(2)={2}^{4}-2(4{)}^{3}=C$

$16-2(64)=C$

$16-128=C$

$-112=C$

Is this what I am being asked to do, or am I going about this the wrong way?

Sorry if my title is a bit confusing. I have the question:

A function f defined for all real numbers satisfies:

${f}^{\prime}(1)=1$, $f(0)=4$, and ${f}^{\u2033}(x)=12{x}^{2}-12x$. Find the value of f(2).

I left out some unimportant parts which I should be capable of doing myself after I understand this bit, my question is am I doing this correctly?

${f}^{\prime}(x)=4{x}^{3}-6{x}^{2}+C$

$4(1{)}^{3}-6(1{)}^{2}+C=1$

$4-6+C=1$

$C=3$

$f(x)={x}^{4}-2{x}^{3}+C$

$f(0)={0}^{4}-2(0{)}^{3}+C=4$

$C=4$

$f(2)={2}^{4}-2(4{)}^{3}=C$

$16-2(64)=C$

$16-128=C$

$-112=C$

Is this what I am being asked to do, or am I going about this the wrong way?

kubistiedt

Beginner2022-05-01Added 17 answers

Step 1

Your first step (finding f′(x)) looks good, but from there you miss a few things. After finding the first C, you have ${f}^{\mathrm{\prime}}(x)=4{x}^{3}-6{x}^{2}+3$

Step 2

Thus, when we integrate, we have $f(x)={x}^{4}-2{x}^{3}+3x+C$ where here C is a different constant than the first one. Now we use $f(0)=4$ to find this C, after which you can plug in $x=2$

Your first step (finding f′(x)) looks good, but from there you miss a few things. After finding the first C, you have ${f}^{\mathrm{\prime}}(x)=4{x}^{3}-6{x}^{2}+3$

Step 2

Thus, when we integrate, we have $f(x)={x}^{4}-2{x}^{3}+3x+C$ where here C is a different constant than the first one. Now we use $f(0)=4$ to find this C, after which you can plug in $x=2$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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