Find a function f(x) such that the point (-1,1) is on the graph of y=f(x), the slope of the tangent line at (-1,1) is 2 and f” (x) = 6x + 4?

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Answer 1 · 2018-05-07T09:17:41

#f(x)=x^3+2x^2+3x+4#

We have #f''(x)=6x+4#

#f'(x)=int(6x+4)dx=3x^2+4x+c_1#

As slope of tangent at #(-1,1)# is #2#,

#f'(-1)=3*(-1)^2+4*(-1)+c_1=2# or #3-4+c_1=2# i.e. #c_1=3#

Hence #f'(x)=3x^2+4x+3#

and #f(x)=int(3x^2+4x+3)dx=x^3+2x^2+3x+c_2#

Now point #(-1,1)# lies on this curve, hence

#f(-1)=(-1)^3+2*(-1)^2+4*(-1)+c_2=1#

or #-1+2-4+c_1=1# i.e. #c_1=1+1-2+4=4#

Hence #f(x)=x^3+2x^2+3x+4#