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Chapter 3: Problem 35

angent lines a. Find the derivative function \(f^{\prime}\) for the following functions \(f\) b. Find an equation of the line tangent to the graph of \(f\) at \((a, f(a))\) for the given value of \(a\). $$f(x)=3 x^{2}+2 x-10 ; a=1$$

Short Answer

Expert verified
Question: Find the equation of the tangent line to the graph of \(f(x)=3x^2+2x-10\) at the point where \(a=1\). Answer: The equation of the tangent line at the point where \(a = 1\) is \(y = 8x - 13\).

Step by step solution

01

a. Find the derivative function \(f'(x)\)

To find the derivative of the function \(f(x)=3x^2 + 2x - 10\), we apply the following differentiation rules: - Power Rule: \((x^n)'= nx^{n-1}\) - Constant Rule: \((c)' = 0\) - Sum Rule: \((g(x) + h(x))' = g'(x) + h'(x)\), for functions \(g(x)\) and \(h(x)\) Applying these rules to \(f(x)\), we get: $$f'(x) = \frac{d}{dx}(3x^2 + 2x - 10) = 3\frac{d}{dx}(x^2) + 2\frac{d}{dx}(x) - \frac{d}{dx}(10)$$ $$f'(x) = 3(2x^{2-1}) + 2(1x^{1-1}) - 0$$ $$f'(x) = 6x + 2$$ Thus, the derivative function \(f'(x)\) is \(6x + 2\).
02

b. Calculate the point \((a,f(a))\)

We are given the value of \(a=1\). To find the coordinates of the point \((a, f(a))\), we substitute the value of \(a\) into the function \(f(x)\): $$f(a) = f(1) = 3(1)^2 + 2(1) - 10 = 3 + 2 -10 = -5$$ So, the point \((a, f(a))\) is \((1, -5)\).
03

b. Find the slope of the tangent

Now that we have the derivative function \(f'(x) = 6x+2\), we substitute the value of \(a=1\) into the derivative function to find the slope of the tangent at the point \((1, -5)\): $$m = f'(a) = f'(1) = 6(1) + 2 = 8$$ So, the slope of the tangent line \(m = 8\).
04

b. Finding the tangent line equation

To find the equation of the tangent line, we now have the point \((1, -5)\) and the slope \(m=8\). We can use the point-slope form of a linear equation, which is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the given point, and \(m\) is the slope. Substituting the known values, we get: $$y - (-5) = 8(x - 1)$$ $$y + 5 = 8x - 8$$ $$y = 8x - 13$$ This is the equation of the line tangent to the graph of \(f(x)\) at \((a, f(a))\), where \(a=1\).

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Key Concepts

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

Derivative Function
In calculus, the derivative function is a fundamental concept that reflects how a function's output value changes as its input value changes. Essentially, the derivative is the rate of change or the slope of the function at any point. To find the derivative of a function, various rules and operations are used, such as the power rule, product rule, and chain rule, to mention a few.

For the exercise we discussed, the derivative function of f(x) = 3x^2 + 2x - 10 is found by applying differentiation rules to each term separately. The resulting derivative function, f'(x) = 6x + 2, gives the slope of the tangent line to the graph of f at any point x. Understanding how to derive this function is important because it allows us to calculate the exact slope at any given point on the curve, which is part of finding the equation of a tangent line.
Power Rule Differentiation
The power rule differentiation is a quick way to differentiate functions of the form x^n where n is any real number. The power rule states that the derivative of x^n with respect to x is nx^(n-1). This rule is part of the backbone of calculus and is used often for finding the slopes of functions at particular points.

In the exercise above, the power rule is applied to the terms 3x^2 and 2x to find f'(x). The constants in a function, like -10 in our example, have a derivative of zero. By applying the power rule, students can confidently derive basic polynomials and build up their skill set for tackling more complex derivatives.
Point-Slope Form
When it comes to writing equations of lines, the point-slope form is incredibly useful, especially when dealing with tangent lines in calculus. This form is given by the equation y - y_1 = m(x - x_1), where (x_1, y_1) is a point on the line and m is the slope of the line.

For our specific problem, the point-slope form was used with the point (1, -5) and the slope 8 to find the equation of the tangent line. This form is versatile and becomes quite handy because it allows for the creation of a line's equation as long as one point on the line and the slope are known.
Slope of a Tangent Line
The slope of a tangent line is a measure of how steep the line is at the point where it touches the function's curve. Calculating this slope involves evaluating the derivative function at the point of tangency. In our exercise, after finding the derivative f'(x), the slope of the tangent at the given point x = 1 is calculated as f'(1) = 8.

This value is not just a number鈥攊t represents the steepness and the direction (whether upward or downward) of the tangent line at that specific point where x = 1. By understanding this concept, students learn not just to calculate the slope, but also to interpret what it implies about the behavior of the function at that point.

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Most popular questions from this chapter

Earth's atmospheric pressure decreases with altitude from a sea level pressure of 1000 millibars (a unit of pressure used by meteorologists). Letting \(z\) be the height above Earth's surface (sea level) in kilometers, the atmospheric pressure is modeled by \(p(z)=1000 e^{-z / 10}\) a. Compute the pressure at the summit of Mt. Everest, which has an elevation of roughly \(10 \mathrm{km} .\) Compare the pressure on Mt. Everest to the pressure at sea level. b. Compute the average change in pressure in the first \(5 \mathrm{km}\) above Earth's surface. c. Compute the rate of change of the pressure at an elevation of \(5 \mathrm{km}\) d. Does \(p^{\prime}(z)\) increase or decrease with \(z ?\) Explain. e. What is the meaning of \(\lim _{z \rightarrow \infty} p(z)=0 ?\)

Find \(d y / d x,\) where \(\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}+x\right)=8 x y^{2}.\)

Suppose \(f\) is differentiable on [-2,2] with \(f^{\prime}(0)=3\) and \(f^{\prime}(1)=5 .\) Let \(g(x)=f(\sin x)\) Evaluate the following expressions. a. \(g^{\prime}(0)\) b. \(g^{\prime}\left(\frac{\pi}{2}\right)\) c. \(g^{\prime}(\pi)\)

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. A mechanical oscillator (such as a mass on a spring or a pendulum) subject to frictional forces satisfies the equation (called a differential equation) $$y^{\prime \prime}(t)+2 y^{\prime}(t)+5 y(t)=0$$ where \(y\) is the displacement of the oscillator from its equilibrium position. Verify by substitution that the function \(y(t)=e^{-t}(\sin 2 t-2 \cos 2 t)\) satisfies this equation.

Let \(f\) and \(g\) be differentiable functions with \(h(x)=f(g(x)) .\) For a given constant \(a,\) let \(u=g(a)\) and \(v=g(x),\) and define $$H(v)=\left\\{\begin{array}{ll} \frac{f(v)-f(u)}{v-u}-f^{\prime}(u) & \text { if } v \neq u \\ 0 & \text { if } v=u \end{array}\right.$$ a. Show that \(\lim _{y \rightarrow u} H(v)=0\) b. For any value of \(u,\) show that $$f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u)$$ c. Show that $$h^{\prime}(a)=\lim _{x \rightarrow a}\left(\left(H(g(x))+f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\right)$$ d. Show that \(h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)\)
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