BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 2.6, Problem 7E
To determine

To find: The equation of the tangent line to the curve at the given point.

Expert Solution

Answer to Problem 7E

The equation of the tangent line to the curve y=x at the point (1, 1) is y=12x+12.

Explanation of Solution

Given:

The equation of the curve is y=x.

The curve passing through the point (1, 1).

Formula used:

The slope of the tangent curve y=f(x) at the point P(a,f(a)) is,

m=limxaf(x)f(a)xa (1)

The equation of the tangent line to the curve y=f(x) at the point (a,f(a)) is,

yf(a)=f(a)(xa) (2)

Difference of square formula: (a2b2)=(a+b)(ab)

Calculation:

Obtain the slope of the tangent line to the parabola at the point (1,1).

Substitute a=1 and f(a)=1 in equation (1),

m=limx1f(x)f(1)x1=limx1(x)1x1=limx1x1x1

Multiply both the numerator and denominator by the conjugate of the numerator.

m=limx1(x1x1×x+1x+1)

Apply the difference of squares formula,

m=limx1((x)2(1)2)(x1)(x+1)=limx1(x1)(x1)(x+1)

Since the limit x approaches 1 but is not equal to 1, cancel the common term (x1)(0) from both the numerator and the denominator,

m=limx11(x+1)=1(1+1)=11+1=12

Thus, the slope of the tangent line to the curve at the point (1, 1) is m=12_.

Obtain the equation of the tangent line.

Since the tangent line to the curve y=f(x) at (a,f(a)) is the line through the point (a,f(a)) whose slope is equal to the derivative of f at a, which is f(a)=12.

Substitute a=1,f(a)=1 and f(a)=12 in equation (2),

(yf(a))=f(a)(xa)y(1)=12(x1)y1=12x12

Isolate y as shown below.

y=12x12+1=12x+12

Thus, the equation of the tangent line is y=12x+12.

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