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

`\displaystyle h(x) = x-(13)/(2)`

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) & = (1)/(2)(2)(3) + (1)/(2)(2)(-3)+(1)/(2)(-3+(-2))(1) \\ \\ & = (1)/(2)(6-6-5) \\ \\ & = -(5)/(2)\end{aligned}`

Note that areas under the x-axis are negative.

Find g'(4):

`\displaystyle \begin{aligned} g'(x) = (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(-(5)/(2)\right) & = (1)(x-(4)) \\ \\ h(x)& = (x-4) - (5)/(2) \\ \\ & = x - (13)/(2)\end{aligned}`

In conclusion, the equation of the tangent line is:

`\displaystyle h(x) = x-(13)/(2)`