/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Find an equation of the line tan... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 77

Find an equation of the line tangent to the graph of \(y=2^{x}+1\) at the point \((0,2)\).

Short Answer

Expert verified
The equation of the line tangent to the graph of \(y=2^x+1\) at the point \((0,2)\) is \(y=\ln{(2)}x+2\).

Step by step solution

01

Find the derivative of the given function

First, we need to find the derivative of the function \(y=2^x+1\). The derivative of a sum is the sum of the derivatives. The derivative of a constant is 0. To find the derivative of \(2^x\), we can use the chain rule which states that \(\frac{d}{dx}(a^u)=a^u \cdot \ln{(a)}\cdot u'\), where \(u\) is a function of x and \(a\) is a constant. So, \[ \frac{dy}{dx} = \frac{d}{dx}(2^x+1) = \frac{d}{dx}(2^x) + \frac{d}{dx}(1) \] \[ \frac{dy}{dx}=2^x\cdot \ln{(2)}\cdot 1 \] The derivative is \(y' = 2^x \cdot \ln{(2)}\).
02

Evaluate the derivative at the given point

Now we need to find the slope of the tangent line at the point \((0, 2)\) by evaluating the derivative at \(x=0\): \[ y'(0) = 2^0 \cdot \ln{(2)} = 1\cdot \ln{(2)} = \ln{(2)} \] The slope is \(\ln{(2)}\).
03

Use the point-slope form

Finally, we can use the point-slope formula to find the equation of the tangent line. The point-slope formula is: \[ y - y_1 = m(x - x_1) \] where \((x_1,y_1)\) is a point the line passes through and \(m\) is the slope. Plugging in our point \((0, 2)\) and the slope \(\ln{(2)}\): \[ y-2=\ln{(2)}(x-0) \]
04

Simplify the equation of the tangent line

Now, we can simplify the equation of the tangent line to get it in the slope-intercept form: \[ y= \ln{(2)}x + 2 \] The equation of the line tangent to the graph of \(y=2^x+1\) at the point \((0,2)\) is \(y=\ln{(2)}x+2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
Understanding the derivative is key when finding the tangent line to a function. A derivative represents the rate at which a function is changing at any given point. In simple terms, it tells us how steep the graph is at that point. Let's consider the function given:
  • Function: \( y = 2^x + 1 \)
  • To differentiate this, we need the derivative of \( 2^x \).
  • The derivative provides the slope of the tangent line at a specific point along the curve.
By taking the derivative, we obtain a new function:
  • Derivative: \( y' = 2^x \cdot \ln{(2)} \).
This tells us how quickly \( y \) changes with respect to \( x \). Once you have this, you can find the specific slope of the tangent by inputting the \( x \)-value of the point you are interested in.
Chain Rule
The chain rule is a fundamental tool in calculus for finding derivatives of composite functions. A composite function is essentially a function within another function. If you encounter a situation where the function's exponent is another function of \( x \), the chain rule comes in handy.
  • The chain rule formula is \( \frac{d}{dx}(a^u) = a^u \cdot \ln(a) \cdot u' \), where \( u \) is a function of \( x \).
In the original problem, when dealing with \( 2^x \), even though \( x \) is not a composite function, this rule simplifies finding the derivative. We treat \( x \) here as \( u \) with its derivative being 1, thus simplifying our process:
  • Use of Chain Rule: \( y' = 2^x \cdot \ln(2) \cdot 1 \).
Using the chain rule ensures accuracy when differentiating exponential functions, especially as they become more complex.
Point-Slope Form
The point-slope form is vital in writing the equation of a line when you know a point on the line and its slope. This form is given by the formula:
  • Formula: \( y - y_1 = m(x - x_1) \)
You use this formula when you need to write an equation of a line in a simple, easy-to-use format with the given slope and point:
  • Point: \((x_1, y_1) = (0, 2) \)
  • Slope: \( m = \ln(2) \)
Substituting these into the formula gives:
  • Initial Equation: \( y - 2 = \ln(2)(x - 0) \)
This equation can then be rearranged to get a useful form for interpreting the results, allowing you to see where the line crosses the \( y \)-axis (the intercept) and how it moves as \( x \) changes.
Exponential Functions
Exponential functions are mathematical expressions involving exponents where the base is a constant and the exponent is a variable. These types of functions grow very quickly, which often results in steep slopes.
  • General Form: \( y = a^x + b \)
  • Growth Nature: Changes in \( x \) result in exponential changes in \( y \).
For the function \( y = 2^x + 1 \) used in the problem, it's important to note:
  • Additional Constant: The +1 shifts the entire curve up by 1 unit.
  • Base of Exponent: The base is 2, which affects the rate of growth; hence, as \( x \) increases, \( y \) increases exponentially.
Understanding exponential functions' behavior allows us to predict changes over the interval and effectively utilize the derivative to find tangent lines because the slope becomes very steep quite quickly.

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Most popular questions from this chapter

Let \(g\) denote the inverse of the function \(f\). (a) Show that the point \((a, b)\) lies on the graph of \(f .\) (b) Find \(g^{\prime}(b)\) $$ f(x)=\left(x^{3}+1\right)^{3} ; \quad(1,8) $$

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