Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.

(a) Use a linear approximation to estimate g(2.9) and g(3.1). g(2.9) ≈ g(3.1) ≈
(b) Are your estimates in part (a) too large or too small? Explain.

Respuesta :

Answer:

a)

[tex]g(2.9) \approx -6.6[/tex]

[tex]g(3.1) \approx -3.4[/tex]

b)

The values are too small since [tex]g''[/tex] is positive for both values of [tex]x[/tex] in. I'm speaking of the [tex]x[/tex] values, 2.9 and 3.1.

Step-by-step explanation:

a)

The point-slope of a line is:

[tex]y-y_1=m(x-x_1)[/tex]

where [tex]m[/tex] is the slope and [tex](x_1,y_1)[/tex] is a point on that line.

We want to find the equation of the tangent line of the curve [tex]g[/tex] at the point [tex](3,-5)[/tex] on [tex]g[/tex].

So we know [tex](x_1,y_1)=(3,-5)[/tex].

To find [tex]m[/tex], we must calculate the derivative of [tex]g[/tex] at [tex]x=3[/tex]:

[tex]m=g'(3)=(3)^2+7=9+7=16[/tex].

So the equation of the tangent line to curve [tex]g[/tex] at [tex](3,-5)[/tex] is:

[tex]y-(-5)=16(x-3)[/tex].

I'm going to solve this for [tex]y[/tex].

[tex]y-(-5)=16(x-3)[/tex]

[tex]y+5=16(x-3)[/tex]

Subtract 5 on both sides:

[tex]y=16(x-3)-5[/tex]

What this means is for values [tex]x[/tex] near [tex]x=3[/tex] is that:

[tex]g(x) \approx 16(x-3)-5[/tex].

Let's evaluate this approximation function for [tex]g(2.9)[/tex].

[tex]g(2.9) \approx 16(2.9-3)-5[/tex]

[tex]g(2.9) \approx 16(-.1)-5[/tex]

[tex]g(2.9) \approx -1.6-5[/tex]

[tex]g(2.9) \approx -6.6[/tex]

Let's evaluate this approximation function for [tex]g(3.1)[/tex].

[tex]g(3.1) \approx 16(3.1-3)-5[/tex]

[tex]g(3.1) \approx 16(.1)-5[/tex]

[tex]g(3.1) \approx 1.6-5[/tex]

[tex]g(3.1) \approx -3.4[/tex]

b) To determine if these are over approximations or under approximations I will require the second derivative.

If [tex]g''[/tex] is positive, then it leads to underestimation (since the curve is concave up at that number).

If [tex]g''[/tex] is negative, then it leads to overestimation (since the curve is concave down at that number).

[tex]g'(x)=x^2+7[/tex]

[tex]g''(x)=2x+0[/tex]

[tex]g''(x)=2x[/tex]

[tex]2x[/tex] is positive for [tex]x>0[/tex].

[tex]2x[/tex] is negative for [tex]x<0[/tex].

That is, [tex]g''(2.9)>0 \text{ and } g''(3.1)>0[/tex].

So [tex]2x[/tex] is positive for both values of [tex]x[/tex] which means that the values we found in part (a) are underestimations.