Calculus Math help

Question

I was really struggling with these two questions

Consider the following equations. In each case suppose that we apply the Intermediate Value Theorem using the interval [0, 1]. (i.e., we take a = 0, b = 1 in the Intermediate Value Theorem.)

(i)  x2 + x − 1  =  0

(ii)  2ex  =  x − 2

(iii)  ln(x+1)  =  5 − 2x For which equations does the Intermediate Value Theorem conclude that there must be a root of the equation in the interval (0, 1)?

(a) If an equation of the tangent line to the curve y = f (x) at the point where a = 2 is y = 6x − 5, find  f (2) and  f '(2).

(b) If the tangent line to y = f (x) at (-5, 3) passes through the point (8, 7), find  f (-5) and  f '(-5).

Answer 1

(i) f (x) = x² + x - 1 is a polynomial so is continuous. f(0) = -1 and f(1) = 1. As f(0) < 0 and f(1) > 0 and f(x) is continuous by the IVT there must be at least one root in the interval (0,1)

(ii) 2 e ^x = x - 2. Rearrange so we have f(x) = 2 e^x - x + 2. f(0) = 3 and f(1) = 6.4, f(0) > 0 and f(1)>0 so we cannot say from the IVT that there is a root in (0,1)

(iii) Rearrange, f(x) = ln (x+1) - 5 + 2x. f(0) = -5 and f(1) = -2.3 so f(0) < 0 and f(1) < 0 so same situation as (ii), we cannot conclude there is a root in (0,1)

(a)

f'(2) is the slope of the tangent line at x = 2. As the tangent line is 6x - 5, f'(2) = 6

The tangent line touches the curve at the point ( 2, f(2)). So to find f(2) we just put 2 in the equation for the tangent line (as the tangent line and the curve intersect at x = 2). So f(2) = 6 x 2 - 5 = 7

(b) Find the equation of the tangent line. It goes through (-5,3) and (8,7).

So the slope will be (7-3) / (8 --5) = 4 / 13 so f'(-5) = 4/13 from argument in (a)

y = 4/13 x + C, passes through (8.7) so 7 = (4/13) x 8 + c so c = 7 - (32/13)

so tangent line is y = (4/13) x + 7 - (32/13)

To find f(-5) substitute x = -5 into this (same argument as in (a))

Mike