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MTH1020 - Analysis of change - S2 2025

MTH1020 - Analysis of change - S2 2025

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Choose the correct statement/s from below.

If f(x) changes concavity at x=a then there is a point of inflection at x=a.

If f(x) has a point of inflection at x=a, then the tangent line crosses the graph of the function at this point.

If f(x) has a point of inflection at x=a, then f'(x)=0 at x=a.

If f(x) has a point of inflection at x=a then x=a must be in the domain of f.

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A function f is a positive function if $f \geq 0$ f \geq 0 for all values in its domain. Similarly, a function is a negative function if $f \leq 0$ f \leq 0 for all values in its domain.

Consider a differentiable function f and choose the two correct statements from the following choices.

If f is a strictly increasing function then its derivative is a positive function.

If f is a strictly increasing function then its derivative is also a strictly increasing function.

If f is a positive function then its derivative is an odd function.

If f is an increasing function then its derivative is a negative function.

If f is an increasing function then its derivative is a positive function.

If f is a negative function then its derivative is also a negative function.

If f is an increasing function then its derivative is an even function.

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Choose from below, the correct response/s.

The constant function, f(x) = -5 is:

strictly decreasing

increasing

decreasing

strictly increasing

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Consider the function $f(x) = \text{Cos}^{-1}(x) + \text{Tan}^{-1}x$ f(x) = \text{Cos}^{-1}(x) + \text{Tan}^{-1}x. Which of the following options below is correct?

$f'(x) = \dfrac{1}{\sqrt{1-x^2}} + \dfrac{1}{1+x^2}$ f'(x) = \dfrac{1}{\sqrt{1-x^2}} + \dfrac{1}{1+x^2}

$f'(x) = \dfrac{1}{\sqrt{1+x^2}} + \dfrac{1}{1-x^2}$ f'(x) = \dfrac{1}{\sqrt{1+x^2}} + \dfrac{1}{1-x^2}

$f'(x) = \dfrac{1}{\sqrt{1-x^2}} - \dfrac{1}{\sqrt{1+x^2}}$ f'(x) = \dfrac{1}{\sqrt{1-x^2}} - \dfrac{1}{\sqrt{1+x^2}}

$f'(x) = -\dfrac{1}{\sqrt{1-x^2}} + \dfrac{1}{1+x^2}$ f'(x) = -\dfrac{1}{\sqrt{1-x^2}} + \dfrac{1}{1+x^2}

$f'(x) = \dfrac{1}{\sqrt{1-x^2}} + \dfrac{1}{\sqrt{1+x^2}}$ f'(x) = \dfrac{1}{\sqrt{1-x^2}} + \dfrac{1}{\sqrt{1+x^2}}

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Consider the function $h(x) = \sin(x) - \cos(x)$ h(x) = \sin(x) - \cos(x). Which of the following options below is correct?

$h'(x) = - \cos(x) + \sin(x)$ h'(x) = - \cos(x) + \sin(x)

$h'(x) = - \cos(x) - \sin(x)$ h'(x) = - \cos(x) - \sin(x)

$h'(x) = \sin(x) + \cos(x)$ h'(x) = \sin(x) + \cos(x)

$h'(x) = \cos(x) - \sin(x)$ h'(x) = \cos(x) - \sin(x)

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Consider the function $g(x) = \ln(x)+x^2$ g(x) = \ln(x)+x^2. Which of the following options below is correct?

$g'(x) = \dfrac{1}{x} + 2x$ g'(x) = \dfrac{1}{x} + 2x

$g'(x) = \dfrac{1}{\ln(x^2)}$ g'(x) = \dfrac{1}{\ln(x^2)}

$g'(x) = \dfrac{1}{\ln(x)} + 2x$ g'(x) = \dfrac{1}{\ln(x)} + 2x

$g'(x) = \dfrac{1}{x}$ g'(x) = \dfrac{1}{x}

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Consider the function $f(x) = 3e^{2x}$ f(x) = 3e^{2x}. Which of the following options below is correct?

$f'(x) = 3e^{2x}$ f'(x) = 3e^{2x}

$f'(x) = 6x\,e^{2x}$ f'(x) = 6x\,e^{2x}

$f'(x) = 6e^{x}$ f'(x) = 6e^{x}

$f'(x) = 6e^{2x}$ f'(x) = 6e^{2x}

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Consider the functions f(x) = x^3 and g(x) = 2x.

Which of the following choices below shows the correct use of the chain rule to find the derivative of the composition function $\big(f \circ g\big)(x)$ \big(f \circ g\big)(x), that is, for $\big(f \circ g\big)'(x)$ \big(f \circ g\big)'(x)?

$\big(f \circ g\big)'(x) = 3x^2\, 2 = 6x^2$ \big(f \circ g\big)'(x) = 3x^2\, 2 = 6x^2

$\big(f \circ g\big)'(x) = 3(2x)^2 \,2 = 24x^2$ \big(f \circ g\big)'(x) = 3(2x)^2 \,2 = 24x^2

$\big(f \circ g\big)'(x) = 3(2x)^2 \,2x = 24x^3$ \big(f \circ g\big)'(x) = 3(2x)^2 \,2x = 24x^3

$\big(f \circ g\big)'(x) = 3(2x^2) \,2 = 12x^2$ \big(f \circ g\big)'(x) = 3(2x^2) \,2 = 12x^2

$\big(f \circ g\big)'(x) = 3x^2\, 2x = 6x^3$ \big(f \circ g\big)'(x) = 3x^2\, 2x = 6x^3

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