# Calculus Bible online by Gill G.S.

By Gill G.S.

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Example text

Part (iii) The slope of the line is given by m= −2 − 4 = −3. 5−3 The equation of this line is y − 4 = −3(x − 3) or y + 2 = −3(x − 5). 2. LINEAR FUNCTION APPROXIMATIONS 63 On solving for y, we get the equation of the line as y = −3x + 13. This line goes through the point (0, 13). The number 13 is called the yintercept. The above equation is called the slope-intercept form of the line. 2 Determine the equations of the lines satisfying the given conditions: (i) slope = 3, passes through (2, 4) (ii) slope = −2, passes through (1, −3) (iii) slope = m, passes through (x1 , y1 ) (iv) passes through (3, 0) and (0, 4) (v) passes through (a, 0) and (0, b) Part (i) If (x, y) is on the line, then we equate the slopes and simplify: 3= y−4 x−2 or y − 4 = 3(x − 2).

This form of the line is called the “point-slope” form of the line.

Sech (x) = = x , for all real x, read as hyperbolic secant of cosh x e + e−x x. 6. csch (x) = 2 1 = x , x = 0, read as hyperbolic cosecant of x. 1 Eliminate quotients and exponents in the following equation by taking the natural logarithm of both sides. 4. 2 Solve the following equation for x: log3 (x4 ) + log3 x3 − 2 log3 x1/2 = 5. Using logarithm properties, we get 4 log3 x + 3 log3 x − log3 x = 5 6 log3 x = 5 5 log3 x = 6 5/6 x = (3) . 3 Solve the following equation for x: 1 ex = . x 1+e 3 On multiplying through, we get 3ex = 1 + ex or 2ex = 1, ex = 1 2 x = ln(1/2) = − ln(2).