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

Baublysiz

Baublysiz

Answered question

2021-11-19

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

Answer & Explanation

Greg Snyder

Greg Snyder

Beginner2021-11-20Added 11 answers

f(x)=3x3
81x+y=0 is tangent to f(x)
slope of tangent,f(x)=ab
f(x)=811
f(x)=81
given f(x)=3x3
3x3=81
x3=27
x=3
substitute in 81x+y=0
243+y=0
y=243
Therefore (x,y)=(-3,243)
we have f(x)=3x3
integrate
f(x)=3x44+c
substitute (-3,243) 243=3(3)44+c
243=2434+c
c=2432434
c=7294
Therefore f(x)=3x44+7294

Thouturs

Thouturs

Beginner2021-11-21Added 12 answers

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
ax4+bx3+cx2+dx+c=0
f(x)=4ax3+3bx2+2cx+d(i)
now since f(x)=3x3(ii)
so from equation i and ii we get
4a=3 , b, c,d=0
a=34
Thus the equation is f(x)=(34)x4+c=0
slope of line 81x+y=0 is
m=181
3x3=1243
x=(1243)13
x=-0.160
Thus equation of the graph is f(x)=(34)x4
-0.160 answer

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