 Baublysiz

2021-11-19

Find a function f such that ${f}^{\prime }\left(x\right)=3{x}^{3}$ and the line 81x+y=0 is tangent to the graph of f. Greg Snyder

${f}^{\prime }\left(x\right)=3{x}^{3}$
81x+y=0 is tangent to f(x)

$⇒{f}^{\prime }\left(x\right)=-\frac{81}{1}$
$⇒{f}^{\prime }\left(x\right)=-81$
given ${f}^{\prime }\left(x\right)=3{x}^{3}$
$⇒3{x}^{3}=-81$
$⇒{x}^{3}=-27$
$⇒x=-3$
substitute in 81x+y=0
$⇒-243+y=0$
$⇒y=243$
Therefore (x,y)=(-3,243)
we have ${f}^{\prime }\left(x\right)=3{x}^{3}$
integrate
$f\left(x\right)=\frac{3{x}^{4}}{4}+c$
substitute (-3,243) $⇒243=\frac{3{\left(-3\right)}^{4}}{4}+c$
$⇒243=\frac{243}{4}+c$
$⇒c=243-\frac{243}{4}$
$⇒c=\frac{729}{4}$
Therefore $f\left(x\right)=\frac{3{x}^{4}}{4}+\frac{729}{4}$ Thouturs

Since the first derivative is of the order 3 thus f(x) must be having order 4 so general equation of a 4 degree equation is
$a{x}^{4}+b{x}^{3}+c{x}^{2}+dx+c=0$
${f}^{\prime }\left(x\right)=4a{x}^{3}+3b{x}^{2}+2cx+d\dots \left(i\right)$
now since ${f}^{\prime }\left(x\right)=3{x}^{3}\dots \left(ii\right)$
so from equation i and ii we get
4a=3 , b, c,d=0
$a=\frac{3}{4}$
Thus the equation is $f\left(x\right)=\left(\frac{3}{4}\right){x}^{4}+c=0$
slope of line $81x+y=0$ is
$m=-\frac{1}{81}$
$3{x}^{3}=-\frac{1}{243}$
$x={\left(-\frac{1}{243}\right)}^{\frac{1}{3}}$
x=-0.160
Thus equation of the graph is $f\left(x\right)=\left(\frac{3}{4}\right){x}^{4}$