What is the equation of the tangent line of f(x)=2x3−1x2 at x=1?

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chicasdelperuszp · Accepted Answer

Explanation:the slope of the tangent line is given by f&#039;(x)differentiate using the quotient ruleGiven f(x)=g(x)h(x) thenf′(x)=h(x)g′(x)−g(x)h′(x)(h(x))2← quotient ruleg(x)=2x3−1⇒g′(x)=6x2h(x)=x2⇒h′(x)=2x⇒f′(x)=6x4−2x(2x3−1)x4=2x4+2xx4⇒f′(1)=41=4← slope of tangent⇒f(1)=11=1⇒(1,1)← point on tangent⇒y−1=4(x−1)← point-slope form⇒y=4x−3← equation of tangent

laiedkn · Accepted Answer

The first step is do determine the slope of the curve at x=1. We will do this by taking the derivative of the function.

f (x) = \frac{2 x^{3} - 1}{x^{2}} = 2 x - \frac{1}{x^{2}}

f (x) = 2 x - x^{- 2}

f^{'} (x) = 2 + 2 x^{- 3}

Now, evaluate the derivative to find the slope of the curve at x=1:

f^{'} (1) = 2 + 2 (1^{- 3}) = 2 + 2

f'(1)=4
Now that we know the slope, we need to know the intercept. We'll figure this out by solving for f(x) and backing out the intercept.

f (1) = \frac{2 {(1)}^{3} - 1}{{(1)}^{3}} = \frac{2 \cdot 1 - 1}{1} = 2 - 1 = 1

f(1)=4x+b
1=4(1)+b=4+b
b=-3
Now we have our equation:
y=4x-3

What is the equation of the tangent line of f(x)=\frac{2x^{3}-1}{x^{2}}

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