Find the equations of the tangent line and normal line to the curve f(x) = x*....

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Find the equations of the tangent line and normal line to the curve f(x) = x*. √x at the point (1, 1). Illustrate the curve and these lines. SOLUTION y- y = The derivative of f(x) = x² √√x = x¹x¹²²2² = = X - = || So the slope of the tangent line at (1, 1) is f (1) = tangent line is y = = X = - 1 x + The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of %/2, that is g. Thus the equation of the normal line is is Therefore an equation of the - (point-slope form) or (slope-intercept form) (slope-intercept form) We graph the curve and its tangent line and normal line in the figure to the left. (point-slope form) or Find the equations of the tangent line and normal line to the curve f(x) = x+. √x at the point (1, 1). Illustrate the curve and these lines. SOLUTION y- y = The derivative of f(x) = x² √√x = x¹x¹²²2² = = X - = || So the slope of the tangent line at (1, 1) is f (1) = tangent line is y = = X = - 1 x + The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of %/2, that is g. Thus the equation of the normal line is is Therefore an equation of the - (point-slope form) or (slope-intercept form) (slope-intercept form) We graph the curve and its tangent line and normal line in the figure to the left. (point-slope form) or Find the equations of the tangent line and normal line to the curve f(x) = x*. √x at the point (1, 1). Illustrate the curve and these lines. SOLUTION y- y = The derivative of f(x) = x² √√x = x¹x¹²²2² = = X - = || So the slope of the tangent line at (1, 1) is f (1) = tangent line is y = = X = - 1 x + The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of %/2, that is g. Thus the equation of the normal line is is Therefore an equation of the - (point-slope form) or (slope-intercept form) (slope-intercept form) We graph the curve and its tangent line and normal line in the figure to the left. (point-slope form) or Find the equations of the tangent line and normal line to the curve f(x) = x+. √x at the point (1, 1). Illustrate the curve and these lines. SOLUTION y- y = The derivative of f(x) = x² √√x = x¹x¹²²2² = = X - = || So the slope of the tangent line at (1, 1) is f (1) = tangent line is y = = X = - 1 x + The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of %/2, that is g. Thus the equation of the normal line is is Therefore an equation of the - (point-slope form) or (slope-intercept form) (slope-intercept form) We graph the curve and its tangent line and normal line in the figure to the left. (point-slope form) or

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To find the equations of the tangent and normal lines to the curve fx x sqrtx at the point 1 1 we fo... View the full answer