What is the equation of the tangent line of f(x)= x^2-3x+(3x^3)/(x-7) at x=2?

1 Answer

The Tangent line equation
179x+25y=188

Explanation:

Given f(x)=x^2-3x+(3x^3)/(x-7) at x=2

let us solve for the point (x_1, y_1) first

f(x)=x^2-3x+(3x^3)/(x-7)

At x=2

f(2)=(2)^2-3(2)+(3(2)^3)/(2-7)

f(2)=4-6+24/(-5)

f(2)=(-10-24)/5

f(2)=-34/5

(x_1, y_1)=(2, -34/5)

Let us compute for the slope by derivatives

f(x)=x^2-3x+(3x^3)/(x-7)

f' (x)=2x-3+((x-7)*9x^2-(3x^3)*1)/(x-7)^2

Slope m=f' (2)=2(2)-3+((2-7)*9(2)^2-(3(2)^3)*1)/(2-7)^2

m=4-3+(-180-24)/25

m=1-204/25=-179/25

The equation of the Tangent line by Point-Slope Form

y-y_1=m(x-x_1)

y-(-34/5)=-179/25(x-2)

y+34/5=-179/25(x-2)

25y+170=-179(x-2)

25y+170=-179x+358

179x+25y=188

Kindly see the graph of f(x)=x^2-3x+(3x^3)/(x-7) and 179x+25y=188

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God bless....I hope the explanation is useful.