Find a function f such that f '(x) = 5x^3 and the line 40x + y = 0

Cabiolab

Cabiolab

Answered question

2021-09-17

Find a function f such that f(x)=5x3 and the line 40x+y=0 is tangent to the graph of f.

Answer & Explanation

Brighton

Brighton

Skilled2021-09-18Added 103 answers

Step 1
We have to find a function  f  such that  f(x)=5x3  and the line  40x+y=0  is tangent to the graph of  f
First, notice that by integraling  f(x)=5x3  we get
f(x)=5x44+C
where C is a integraling constant .
We have to find the value of C.
Now notice that the slope of any tangent to the graph of f(x) is given by f(x)
Also, note that the slope of the line y=40x is given by -40
Thus in order to the line 40x+y=0 become a tangent to the graph of f(x) We must have a value for x that allows for
f(x)=40
Thus we have solve
5x3=40
This provides
x3=8x=2   [Considering only real cubic roots of -8]
Thus, at x=-2, the line 40x+y=0 become a tangent to the graph of f(x)
Step 2
As x=2,f(x) touch the line y=40x hence, we achieve
f(2)=40×(2)=80
Again pluggin x=-2 on the expression f(x)=5x44+C, we get
80=5×(2)44+CC=8020=60
Therefore, the necessary function is
f(x)=5x44+60
Answer
f(x)=5x44+60

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