Find a function f such that f′(x)=3x3 and the line 81x+y=0 is tangent to the graph of f.

Question

Greg Snyder · Accepted Answer

f′(x)=3x381x+y=0 is tangent to f(x)⇒slope of tangent,f′(x)=−ab⇒f′(x)=−811⇒f′(x)=−81given f′(x)=3x3⇒3x3=−81⇒x3=−27⇒x=−3substitute in 81x+y=0⇒−243+y=0⇒y=243Therefore (x,y)=(-3,243)we have f′(x)=3x3integratef(x)=3x44+csubstitute (-3,243) ⇒243=3(−3)44+c⇒243=2434+c⇒c=243−2434⇒c=7294Therefore f(x)=3x44+7294

Thouturs · Accepted Answer

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
$a x^{4} + b x^{3} + c x^{2} + d x + c = 0$
$f^{'} (x) = 4 a x^{3} + 3 b x^{2} + 2 c x + d \dots (i)$
now since $f^{'} (x) = 3 x^{3} \dots (i i)$
so from equation i and ii we get
4a=3 , b, c,d=0
$a = \frac{3}{4}$
Thus the equation is $f (x) = (\frac{3}{4}) x^{4} + c = 0$
slope of line $81 x + y = 0$ is
$m = - \frac{1}{81}$
$3 x^{3} = - \frac{1}{243}$
$x = {(- \frac{1}{243})}^{\frac{1}{3}}$
x=-0.160
Thus equation of the graph is $f (x) = (\frac{3}{4}) x^{4}$
-0.160 answer

Find a function f such that f'(x)=3x^3 and the line

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