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Thank you for taking my question, I tried to add the necessary brackets so its easy to read.

Given f(x)=x^2/(3x+2) Write an equation of the tangent line to f(x) at x=1

Firstly, let's discover the gradient function by differentiating f(x) wrt x which would involve using the quotient rule:

f'(x) = [(3x+2)*(2x)- (x^2)*(3)]/(3x+2)^2

= (3x^2 +4x)/(3x+2)^2

At x=1, y=f(1)= 1^2/(3+2)=1/5

Value of gradient at x=1 is given by f'(1)= (3+4)/(3+2)^2 =7/25

We can thus proceed to construct the equation of the tangent by employing the template

y-y1= m (x-x1)

Substituting x1=1, y1=1/5 and m=7/25, we have

y-1/5 = 7/25 (x-1)

Multiplying both sides by 25 gives

25y-5=7(x-1)

Arranging the terms gives 25y+2=7x (shown)

Hope this helps. Peace.

Calculus

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