How do you find the slope of a tangent line to the graph of the function #f(x) = x^2/(x+1)# at (2, 4/3)?

Question

Emily Ammerman · Accepted Answer

#"slope "=8/9#

#color(orange)"Reminder"#

The slope of the tangent at x = a on f(x) is f'(a)

To differentiate f(x) use the #color(blue)"quotient rule"#

#"given " f(x)=(g(x))/(h(x))" then "#

#color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(2/2)|)))#

#"here " g(x)=x^2rArrg'(x)=2x#

#"and " h(x)=x+1rArrh'(x)=1#

#rArrf'(x)=((x+1).2x-x^2 .1)/(x+1)^2=(2x^2+2x-x^2)/(x+1)^2#

#=(x^2+2x)/(x+1)^2#

#rArrf'(2)=(4+4)/3^2=8/9#

Luke Alvarez · Answer

undefined To find the slope of a tangent line to the graph of a function at a specific point, you can use the derivative of the function.

To find the derivative of f(x) = x^2/(x+1), you can use the quotient rule. The quotient rule states that if you have a function in the form f(x) = g(x)/h(x), then the derivative of f(x) is given by [g'(x)h(x) - g(x)h'(x)] / [h(x)]^2.

Applying the quotient rule to f(x) = x^2/(x+1), we have:

f'(x) = [(2x)(x+1) - (x^2)(1)] / [(x+1)^2]

Simplifying this expression, we get:

f'(x) = (2x^2 + 2x - x^2) / (x^2 + 2x + 1)

Combining like terms, we have:

f'(x) = (x^2 + 2x) / (x^2 + 2x + 1)

Now, to find the slope of the tangent line at the point (2, 4/3), we substitute x = 2 into the derivative expression:

f'(2) = (2^2 + 2(2)) / (2^2 + 2(2) + 1)

Simplifying this expression, we get:

f'(2) = (4 + 4) / (4 + 4 + 1)

f'(2) = 8 / 9

Therefore, the slope of the tangent line to the graph of f(x) = x^2/(x+1) at the point (2, 4/3) is 8/9.