We must find the equation of the tangent line for the function g(x) at the point x = 5, given:
[tex]g(x)=\int ^x_0f(t)dt.[/tex]The general equation of the tangent line is:
[tex]y=m\cdot x+b\text{.}[/tex]where:
• m is the slope,
,• b is the y-intercept.
1) Slope of the line
To find the equation of the line, first, we must find the slope of the line at x = 5, which is given by the derivative of the function g(x) at x = 5.
The Fundamental Theorem of Calculus says that:
[tex]g^{\prime}(x)=\frac{d}{dx}\int ^x_0f(t)dt=f(x)\text{.}[/tex]Using the graph of the function f, we get:
[tex]g^{\prime}(5)=f(5)=-1.[/tex]So the slope of the line tangent to g(x) at the point x = -1 is:
[tex]m=g^{\prime}(5)=-1.[/tex]So the equation of the line has the form:
[tex]y=h(x)=-x+b\text{.}[/tex]We must find the value of b.
2) y-intercept of the line
If the line is tangent to g(x) at the point with x = 5, the line must have the same value as g(x) at x = 5:
[tex]y=h(5)=-5+b=g(5)\text{.}[/tex]We must compute the value of g(5), which is given by:
[tex]g(5)=\int ^5_0f(t)dt\text{.}[/tex]So the value of g(5) is the area under the curve f(t). Summing the different contributions, we get:
[tex]g(5)=\int ^5_0f(t)dt=1+0.5-0.5-1.5-1.5=-2.[/tex]Replacing the result in the equation above, we get:
[tex]\begin{gathered} -5+b=g(5), \\ -5+b=-2, \\ b=-2+5, \\ b=3. \end{gathered}[/tex]Using the value m = -1 and b = 3, the equation of the tangent line is:
[tex]y=-x+3.[/tex]Answer
[tex]y=-x+3[/tex]