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How do you find the slope of the graph #f(x)=-1/2+7/5x^3# at (0,-1/2)?

Answer 1

#color(green)(0)#

The slope of a function #f(x)# for a non-specific point #x# is given by the derivative of the function #f(x)#.

For the case #f(x)=-1/2+7/5x^3#

the derivative (using the exponent rule) is
#color(white)("XXX")f'(x)=3xx7/5x^(3-1) = 21/5x^2#

At the point #(0,-1/2)#
#color(white)("XXX")x=0# (and #y=-1/2#, but that's irrelevant).
so the slope is
#color(white)("XXX")f'(0)=21/5 * (0)^2 =0#

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Answer 2

Paisley Johnson

To find the slope of the graph ( f(x) = -\frac{1}{2} + \frac{7}{5}x^3 ) at the point ((0, -\frac{1}{2})), you need to find the derivative of the function ( f(x) ) with respect to ( x ) and then evaluate it at ( x = 0 ).

The derivative of ( f(x) ) with respect to ( x ) is ( f'(x) = \frac{d}{dx}(-\frac{1}{2} + \frac{7}{5}x^3) = \frac{21}{5}x^2 ).

Now, plug in ( x = 0 ) into ( f'(x) ) to find the slope at the point ( (0, -\frac{1}{2}) ).

( f'(0) = \frac{21}{5} \times 0^2 = 0 ).

So, the slope of the graph ( f(x) = -\frac{1}{2} + \frac{7}{5}x^3 ) at the point ( (0, -\frac{1}{2}) ) is ( 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.