Question

Let f be the continuous function defined on [-6,6]. Let g be the function given by g(x) = $\int\limits^x_{-1} {f(t)} \, dt$ . Write an equation for the tangent to the graph of g at x=4

Answer 1

Answer:

$\displaystyle h(x) = x-\frac{13}{2}$

Step-by-step explanation:

We are given that:

$\displaystyle g(x) = \int_{-1}^x f(t)\, dt$

And we want to write the equation for the tangent to the graph of g at x = 4.

First, determine g(4):

$\displaystyle \begin{aligned} g(4) & = \int_{-1}^{(4)} f(t)\, dt \end{aligned}$

This is equivalent to the area of f from -1 to 4:

$\displaystyle \begin{aligned} g(4) & = \frac{1}{2}(2)(3) + \frac{1}{2}(2)(-3)+\frac{1}{2}(-3+(-2))(1) \\ \\ & = \frac{1}{2}(6-6-5) \\ \\ & = -\frac{5}{2}\end{aligned}$

Note that areas under the x-axis are negative.

Find g'(4):

$\displaystyle \begin{aligned} g'(x) = \frac{d}{dx}\left[ \int_{-1}^x f(t)\, dt\right]\end{aligned}$

By the Fundamental Theorem of Calculus:
$\displaystyle \begin{aligned} g'(x) & = f'(x) \\ \\ g'(4) & = f'(4) \\ \\ & =1 \end{aligned}$

Note f is a line for 3 < x < 6 with a slope of 1.

Therefore:
$\displaystyle \begin{aligned} h(x) -g(x_1) & = g'(x_1)(x-x_1) \\ \\ h(x) - \left(-\frac{5}{2}\right) & = (1)(x-(4)) \\ \\ h(x)& = (x-4) - \frac{5}{2} \\ \\ & = x - \frac{13}{2}\end{aligned}$

In conclusion, the equation of the tangent line is:
$\displaystyle h(x) = x-\frac{13}{2}$