Tag: Tangents

Find the tangent line to the graph of a cubic at a particular point

by RoRi

September 17, 2015

Consider the cubic curve

$f(x) = x - x^3.$

Find values for $m$ and $b$ such that the line $y = mx + b$ is tangent to the graph of $f$ at the point $(-1,0)$ .

There is a second line passing through the point $(-1,0)$ which is tangent to the graph of $f$ at the point $(a,c)$ . Find the values of $a$ and $c$ .

First, we sketch the graph of $f$ on the interval $[-2,2]$ :

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We compute the derivative of $f$ ,

$f'(x) = 1 - 3x^2.$

So, the slope of the tangent line at the point $(-1,0)$ is

$f'(-1) = 1 - 3(-1)^2 = -2 \quad \implies \quad m = -2.$

Then, since $y = mx + b$ is the tangent line to the curve at $(-1,0)$ it must be on the curve at this point, so

$y(-1) = 0 \quad \implies \quad (-2)(-1) + b = 0 \quad \implies \quad b = -2.$

Therefore, the line is $y = -2x -2$ , i.e., $m = b = -2$ .

Next, if there is another line, say $y_1 = m_1 x + b_1$ , tangent to $f$ at a point $(a,c)$ we know it has slope given by

$f'(a) = 1 - 3a^2 = m_1.$

Since it must pass through the point $(-1,0)$ (by hypothesis) we must have

$y_1(-1) = 0 \quad \implies \quad (1-3a^2)(-1) + b_1 = 0 \quad \implies \quad b_1 = 1 -3a^2.$

So the line $y_1$ is of the form,

$y_1 = (1 - 3a^2)x + (1-3a^2).$

Finally, since the point $(a,c)$ is on both this line $y_1$ and the curve $f$ we have,

$\begin{align*} f(a) &= c & \text{and} && y_1(a) &= c \\ a-a^3 &=c & \text{and} && (1-3a^2)(a) + (1-3a^2) &=c. \end{align*}$

Since these are both equal to $c$ , they must be equal to each other,

$\begin{align*} &&a - a^3 &= (1-3a^2)(a) + (1-3a^2) \\ \implies && 2a^3 + 3a^2 - 1 &= 0 \\ \implies && a &= \frac{1}{2}. \end{align*}$

Since $a - a^3 = c$ we then have $c = \frac{3}{8}$ .

Thus, $a = \frac{1}{2}$ , and $c = \frac{3}{8}$ are the requested values.

Find the point at which a given line is tangent to a given curve

by RoRi

September 17, 2015

Consider the curve given by the cubic equation

$y = x^3 - 6x^2 + 8x.$

Show that the line $y = -x$ is tangent to this curve and find the point of tangency. Determine if this line intersects the curve anywhere else.

First, we compute the derivative of the curve,

$y' = 3x^2 - 12x + 8.$

For the line $y = -x$ to be tangent to the curve $y = x^3 - 6x^2 + 8x$ it must be at a point $x$ such that $y' = -1$ (since this line has slope $-1$ , the derivative of the curve must be $-1$ ). So,

$y' = -1 \quad \implies \quad 3x^2 - 12x + 8 = -1 \quad \implies \quad x = \{ 1, 3\}.$

Next, for the line to be tangent to the curve the point must actually be on the curve. So, we test out the two possible values of $x$ .

If $x = 1$ , we have $y(1) = 1 - 6 + 8 = 3 \neq -1$ . So $y = -x$ is not tangent to the curve at $x = 1$ .

If $x = 3$ , we have $y(3) = 27 - 9(6) + 24 = -3 = -x$ ; hence, $y = -x$ is tangent to the curve at $(3,-3)$ .

This tangent line also intersects the curve at $(0,0)$ .

Find values of constants so two polynomials intersect with the same slope at a point

by RoRi

September 17, 2015

Let $f$ and $g$ be the polynomials:

$f(x) = x^2 + ax + b, \qquad g(x) = x^3 - c.$

Find values of $a,b,c$ such that $f$ and $g$ intersect at the point $(1,2)$ and have the same tangent line at this point.

Since we want $f$ and $g$ to intersect at the point $(1,2)$ we must have $f(1) = g(1) = 2$ . This gives us the equations,

$f(1) = 2 \quad \implies \quad 1 + a + b = 2,$

and,

$g(1) = 2 \quad \implies \quad 1 - c = 2 \quad \implies \quad c = -1.$

Next, we compute the derivatives so that we can find the slope of the tangent lines at this point,

$f'(x) = 2x + a, \qquad g'(x) = 3x^2.$

Since these must be the same at the point $(1,2)$ we have

$f'(1) = g'(1) \quad \implies \quad 2 + a = 3 \quad \implies \quad a = 1.$

Then from above we know $1+a+b = 2$ ; thus, $b = 0$ .

Therefore, the values of the constants are

$a = 1, \qquad b = 0, \qquad c = -1.$

Find constants so a quadratic has a given tangent at a particular point

by RoRi

September 17, 2015

Let $f$ be the quadratic function

$f(x) = x^2 + ax + b \qquad a,b \text{ constants in } \mathbb{R}.$

Find values for $a$ and $b$ such that the graph of $f$ has a tangent line given by $y = 2x$ at the point $(2,4)$ .

First, we compute the derivative

$f'(x) = 2x + a.$

So, if the line $y = 2x$ is tangent to $f$ at the point $(2,4)$ we must have

$f'(2) = 2 \quad \implies \quad 4+a = 2 \quad \implies \quad a = -2.$

Next, the point $(2,4)$ must be on the graph of $f$ , i.e., we must have $f(2) = 4$ . Thus,

$f(2) = 4 \quad \implies \quad 4 + (-2)2 + b = 4 \quad \implies \quad b = 4.$

So, the values are $a = -2$ and $b = 4$ .

Find the points at which the given function has slope zero

by RoRi

September 16, 2015

Consider the function

$f(x) = x + \sin x.$

Find the points $x$ at which the graph of $f$ at $(x,f(x))$ has slope zero.

First, the derivative is given by

$f'(x) = 1 + \cos x.$

Then, the requirement that the graph of $f$ has slope zero at $(x, f(x))$ is asking for the points at which $f'(x) = 0$ . So, we solve,

$f'(x) = 0 \ \implies \ 1 + \cos x = 0 \ \implies \ \cos x = -1 \ \implies \ x = (2n+1) \pi \quad n \in \mathbb{Z}.$

Find points of a function at which the tangent line has specified values

by RoRi

September 16, 2015

Consider the function

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

Find the points at which the slope of $f$ is

The derivative is given by

$f'(x) = 2x^2 + x - 1 = (2x-1)(x+1).$

Then,

The slope is 0 means
$f'(x) = 0 \ \implies \ (2x-1)(x+1) = 0 \ \implies \ x = \left\{ -1, \frac{1}{2} \right\}.$
The slope is -1 means
$f'(x) = -1 \ \implies \ 2x^2 + x - 1 = -1 \ \implies \ 2x^2 + x = 0 \ \implies \ \left\{ 0, -\frac{1}{2} \right\}.$
The slope is 5 means
$f'(x) = 5 \ \implies \ 2x^2 + x - 1 = 5 \ \implies \ 2x^2 + x - 6 = 0 \ \implies \ x = \left\{ -2, \frac{3}{2} \right\}.$